# Translation:Ayil Meshulash/Discourse 4

**This chapter contains sections 81 to 139**

##### Section 81Edit

The sine is half of a chord, whether great or small. The sine of an arc is then half of the chord of an arc twice as large, like so:

Here the sine of arc BC is equal to the line BE, which is half of line BD, which itself is the chord of arc BCD. This arc BCD is twice the size or arc BC.

If one wishes to know the sine of an arc one should draw a radius line from one side of the of the arc to the circle's center and draw a line meeting this radius line forming a right angle. This line is equal to the arc's sine line. In the above example, lines CF and BE are drawn from arc BC, meeting at a right angle.

If the arc is greater than a quarter or the circle [for example, arc BH] then a diameter line is drawn across the circle from one end of the arc [CH] and the line drawn at right angles from the other end of the arc [BD] is the sine.

##### Section 82Edit

The line of the sine value is shared by both the arc and the remainder of the arc above the horizontal, which combined are 180 degrees. In the example above the sine line BE partners arc BC with arc BH. The same is true of the angle it creates and its remainder. In the above example angle BFC and angle BFH together equal 180 degrees. In relation to the sine the arc and angles have the same degree measure.

##### Section 83Edit

The greater the arc or the angle the greater the sine line will be. This rule continues until the arc reaches a quarter or the circle or the angle becomes a 90 degree angle. From that point on if the arc continues to grow the sine line is reduced in length. Therefore, the greatest sine line to be had is the one at half the diameter of the circle, which is the line for a arc measuring a quarter of the circle or an angle that is 90 degrees, as is the line AF above, which is the sine line for arc AC and AH as well as angles AFC and AFH. The sine measurement in all its possible forms is generally divided into 100,000 discrete parts, [or 5 decimal places]. These discrete units are used to measure any arc to determine how many units of this total is the length of the sine line. Through this method a chart of all possible sine lines lengths for each degree of an arc can be measured. Then, if one knows how many degrees are in an angle or an arc he can refer to the table and look up the degree measure and find its corresponding sine line length. Conversely, if one knows the length of a sine line and wishes to know the amount of degrees in the corresponding angle or arc he need only look up the sine value to find its degree measure. If one cannot locate his specific sine line length or degree measure he can locate the next higher value and the previous value, and create a proportion of the fractional units [of the measurement he does know] that are additional to the lower value on the table in comparison to one full unit, and that proportion would be same for the value he is searching for in the corresponding column [for the measurement that he desires to know] compared to its previous and subsequent values.

##### Section 84Edit

The cosine is the length of the horizontal line that is needed to reach a sine line at the 90 degree mark. For example, in the figure above line BG is the cosine to sine line BE, since arc AB added to arc BC is 90 degrees. If the arc extends past 90 degrees the cosine line is the horizontal line extending past the 90 degree point, as line BG in arc BAH above.

##### Section 85Edit

A line proceeding from the center that meets the sine line at right angles is also a cosine line, as seen in line FE above, which is equivalent to line GB. The proof of this was stated in section 22.

##### Section 86Edit

The squares of the sine and cosine lines are equal to the square of the hypotenuse [A^2 + B^2 = C^2]. The proof of this is that hypotenuse BF, which is equivalent to the fullest possible sine line [the radius line at the center, see section 7], is also equal to the squares of lines BG and GF, as was stated in section 75. Since line GF is also equal to line BE, which is the sine line of arc BC, as was written in a previous section, and line BG is the cosine of the sine line BE, the sum of the squares of the sine and cosine equal the square of the hypotenuse.

##### Section 87Edit

As are the proportions that exist between the sines of the angles of a triangle so too are the proportions that exist between the lengths of the sides of a triangle that are opposite each of those respective angles. For example:

The proportion of angle BCD to angle CDB is equal to the proportion of line BD to line BC. The same is true for all the angles and sides. The proof of this is to draw line DC, which is equal in length to line BD, and then draw a circle around this line with a center at point D. Now line BD becomes half the diameter, creating the right triangle BCD, which has the sine line BC that relates to arc BA. In addition, line CD becomes the cosine of angle BDC, as was shown in section 85, and angle CBD is the completion of angle BDC [to 90 degrees], as was shown in section 24. If this is so, line CD is the sine of angle CBD when the triangle under consideration is a right triangle.

##### Section 88Edit

If the triangle under consideration is acute then one should draw a right angle line from the base until point A, like so:

Here the proportion of angle ACB to line AB is the same as the proportion of angle ABC to line AC, as was shown in the previous section. In addition, the proportion of angle ACD to line AD is the same as the proportion of angle ADC to line AC. Now, as angle ACB and ACD are equal, since they are both right angles, and line AC is shared between the two, the remaining proportions of AB:ADC and AD:ABC are therefore equal to each other.

##### Section 89Edit

If the triangle is obtuse the right angle line should be drawn outside the triangle, like so:

In this triangle, the proportion of angle ADC to line AC is equal to the proportion of angle ACD to line AD. So too, the proportion of angle ADB to line AB is equal to the proportion of angle ABD to line AD. If so, just as the proportion of angle ACB to line AB is, so is the proportion of angle ABC to line AC. Furthermore, as is the proportion of an angle to the side opposite it, so is the proportion of the other angles to the sides that are opposite them. Similarly, as is the proportion of the side to the angle opposite it, so is the proportion of the other sides to the angles opposite them.

##### Section 90Edit

At this point it is clear that if one has two triangles that share the same angles the proportion of the lengths of the sides within triangle A to each other is equal to the proportion of the sides of triangle B to each other. Take, for example, these two triangles:

If angle BAC of triangle ABC is equal to angle EDF of triangle DEF, and angle ABC is equal to angle DEF, the proportion of side AB to side BC is equal to the proportion of side DE to side EF. This is true for the remaining angles and sides of these triangles. In addition, just as the proportion of a side of one triangle is equal to its respective partner in the other triangle, the same would be true for all the sides. For example, the proportion of side AB to DE is the same as the proportion of side BC to side EF, and the rest would follow the same pattern.

##### Section 91Edit

From here it is understood that if a line parallel to one of the sides of the triangle was placed inside the triangle, creating a smaller triangle within the larger one, the proportion of the each of sides of the smaller triangle to its other sides is equal to the proportion of the equivalent sides of the larger triangle to its other sides.

##### Section 92Edit

Therefore, if one draws a triangle above the hypotenuse of a triangle in the reverse orientation, both sharing one hypotenuse, and both being right triangles, like so,

the proportion of the sides of triangle BAC are equal to the proportion of the sides of triangle BDE, since angle BAC is equal to angle BDE, both being right angles, and angle ABC is equal to angle BED, as was explained in section 62, it then follows that angle ACB is equal to angle EBD, and the proportion of the sides of each triangle are the identical.

##### Section 93Edit

When considering a triangle, if one knows two of the angles the third can then be determined, for it is the remainder to [180 degrees], as is shown in section 63.

##### Section 94Edit

It is also evident regarding a right triangle that if only one of the angles is known the other can be determined, since the ninety degree angle is already known as an intrinsic property of a right triangle.

##### Section 95Edit

Similarly, in a triangle with two equal length sides, only one angle is necessary to know, (since the other two are equal and add up to 180 degrees.)

##### Section 96Edit

It is also clear that one does not need to know any of the angles of an equilateral triangle or a right triangle with two equal length sides, since all their angles can be determined implicitly.

##### Section 97Edit

If one has a square and draws diagonal lines from each of the two top corners to the bottom corners, four right triangles are created, like so:

The proof is that angle AIB is a right angle due to the fact that the lengths of lines AB and AC are equal, and the lengths of lines BI and CI are equal. Since this is so, and line AI is shared by triangles AIB and AIC, the angles AIB and AIC must be equal to each other, and as right angles that total 180 degrees. If this so, all the angles must be right angles, including angles BID and CID.

##### Section 98Edit

Though one may know the sine line measurement of a certain angle that does mean that he knows whether the angle that gave rise to it is acute or obtuse, as a sine line is the same for an angle and its supplementary angle [that total together to 180 degrees], as was shown in section 82. There are, however, ten methods to determine the correct angle that gave rise to the line:

##### Section 99Edit

The first is as follows:

If the triangle is known to be a right triangle or obtuse then the remaining angle is acute, and was shown in section 64.

##### Section 100Edit

Method 2)

If the unknown angle is equal to one of the other three angles they must both be acute angles, for it is impossible for them to both be obtuse, as we have previously shown. This is certainly true if all three of the angles are equal.

##### Section 101Edit

Method 3)

If there is another angle in the triangle that is known to be larger than the angle under consideration then the angle is an acute angle. This will be true whether the larger angle is obtuse (as we have shown) or acute, since we know the angle under consideration to be smaller.

##### Section 102Edit

Method 4)

If the two other angles together total more than 90 degrees then the angle under consideration is acute, for it is the remainder, as has been shown in section 63.

##### Section 103Edit

Method 5)

If the measure of the sine line of the angle under consideration is equal to the measure of the sine line of one of the other angles they both must be acute, for it is impossible for them both to be obtuse, as we have shown; and it is likewise impossible for one to be acute and the other obtuse, for if this were so these angles would be like an acute and obtuse angle that share one sine line, the two of which already total 180 degrees without the inclusion of the third angle of the triangle, and all three angles of a triangle must total no more than 180 degrees (as was written in section 63).

##### Section 104Edit

Method 6)

If the measure of the sine line of another angle is greater than the measure of the sine line of this angle then this angle must be acute. For example, in this triangle:

where angle BAD is greater than angle ABD, angle ABD must be acute for the following reason: If angle BAD is obtuse then angle ABD must be acute, and if angle BAD is acute then angle ABD must also be acute, for were it to be obtuse then its remainder, angle DBC, would have to be acute, rendering it smaller than angle BAD, which in turn would cause the total degrees of angles ABD and BAD to be more than 180 degrees, which is impossible (as we have shown).

##### Section 105Edit

Method 7)

If the length of the sine lines of the two other angles are both greater than the cosine lengths of each of those two angles those angles must exceed 90 degrees when added together, and the remaining angle must therefore be acute (as was shown in section 102).

##### Section 106Edit

Method 8)

If the sides that are opposite the angle under consideration are equal in length then the angle is acute, since it was already explained that a triangle with two equal length sides must have two angles that are the same. This is certainly true with an equilateral triangle, where all the angles are acute.

##### Section 107Edit

Method 9)

If one side is longer than the side that is opposite the angle under consideration this angle must be acute. This is due to the fact that the proportion of the sides to each other is equal to the proportion of the sine lengths of the angles opposite these sides to each other. Therefore, if the length of the sine line of the angle opposite the longer side is greater than the length of the sine line of the angle under consideration that would make this angle acute (as was written in section 104).

##### Section 108Edit

Method 10)

If the combined area of squares constructed from the two sides that are not opposite the angle under consideration is greater than the area of a square constructed from the side opposite the angle under consideration then this angle will be acute. The reason is that the larger an angle is the larger the side opposite it is. It is further explained in section 75 that a square constructed from the hypotenuse of a right triangle contains an area equal to squares constructed from the horizontal and vertical sides together. If so, in an obtuse triangle the side opposite the obtuse angle would have a square that contains more than the combined squares constructed from the other two sides.

##### Section 109Edit

If one knows three quantities in any triangle the remaining quantities can be deduced. Note that a triangle has six distinct quantities, three angles and three sides. Furthermore, there are four ways to know three quantities in a triangle: 1) three known angles, 2) three known sides, 3) two known sides and one known angle, 4) two known angles and one known side. One should also know that when three angles are known the length of the sides cannot be deduced, for only their proportion to each other can be determined.

##### Section 110Edit

In a right triangle there are only three necessary arrangements of quantities. Since the right angle is known only two other quantities need be known: another angle and side or two sides [or two angles which is not mentioned]. In each of these arrangements there are three possible combinations: 1) An angle and the base or the vertical or the hypotenuse. 2) Two sides - the vertical and the base or the vertical and hypotenuse or the base and hypotenuse. Each of these can be used to determine the other properties.

##### Section 111Edit

If one knows the vertical and the base and wishes to know the hypotenuse one needs only to square the base and the hypotenuse and then sum the results. The square root of this total is the measure of the hypotenuse, as was explained in section 75.

##### Section 112Edit

If one wishes to know one of the angles, using the following triangle as an example, where the sides AB and BC are known, and the angle BAC is unknown, the proportion of AC to BC is the same proportion of the sine of angle ABC to the sine of angle BAC. [Then a sine table can be used to find the angle's degrees].

##### Section 113Edit

If afterward one wishes to know the angle BCA one should subtract angle BAC from 90 degrees and the remainder will be the degrees in angle BCA. The opposite is true as well, as the amount of degrees that need to be added to angle BCA to reach 90 degrees is the measurement of angle BAC.

##### Section 114Edit

If the vertical and the hypotenuse are known, and the base is unknown the hypotenuse can be squared and then the square of the vertical can be subtracted from it. The square root of the remainder is the length of the base. The proof of this is also based on the explanation in section 75.

##### Section 115Edit

Following this, one who wishes to know the angles should proceed according to the instructions in section 112.

##### Section 116Edit

Another approach would be to estimate the proportion of side AC in relation to the sine of the angle opposite, for that proportion would be equal to the proportion of the side AB, which is the vertical, in relation to angle ACB. Angle BAC would then be the completion of angle ACB [to 90 degrees]. With that information one can determine that the proportion of the sine of the angle opposite line AC, which would equal the proportion of angle BAC to side BC, which is the base. **[IS THIS CORRECT?]**

##### Section 117Edit

If the base and hypotenuse are known one could use the same two approaches outlined above for a known vertical and hypotenuse.

##### Section 118Edit

If the angle at point A is known there is no need to calculate the angle at point C, as that must be the remainder that would total 90 degrees. If the height of the vertical is known, one can use the fact that as is the proportion of angle BCA to side AB so is the proportions of all the angles to the sides opposite [to determine lengths of sides]. Another approach would be to determine the length of one other side besides the vertical and the remaining side would be known by the squaring method previously mentioned. Also, if the angle and the base or hypotenuse were known the same method as used for the known vertical can be applied.

##### Section 119Edit

If the base is known and the sum of the vertical and hypotenuse is known, though neither of their lengths are known, the base should be squared and divided by the combined length of the hypotenuse and vertical. When the remainder is subtracted from the combined hypotenuse and vertical, half of that result will be the length of the vertical, while the remaining amount [the remainder and the other half of the result] is the length of the hypotenuse. For example, if the base is 4, and the hypotenuse and vertical are 8 in sum: 4 squared 16. 16 divided by 8 equals 2. 2 subtracted from 8 equals 6. Half of 6 is 3. That is the length of the vertical. The length of the hypotenuse then equals 5. The proof is as follows: If the base and vertical are squared individually, the area of both these squares together equals 25. If the hypotenuse is squared its square also has an area of 25. This true because of the algebraic rule that when a small value is subtracted from a large value, and the remainder is squared, it is equal to the squares of the large and small values less two times the result of multiplying the small by the large value [(large - small)^2 = (large^2 + small^2) - (2 * (small*large))]. For example, (5 less 2) times (5 less 2) equals 9. If the 5 and 2 are squared that equals 29. 29 less (5 times 4), which equals 20, is also 9.

I am also able to express this proof in another manner, like so:

Side CI is the large side, and a square constructed from it would be ACGI. Side HI is the short side, and its square is HIJK. Line GH is the difference between the small and large sides and its square is DEGH. If the area of square HIJK is added to square ACGI, and the area of square ACDF, which the short side times the large, is removed, and finally if square EJFK is removed as well, as it is the second instance of the multiplication of the small and large sides, what remains now is the square DEGH.

So, if in the previous example the square of the hypotenuse is 8 by 8, this, added to the square of the vertical, less 2 times the product of 8 and the unknown length of the vertical is also the same as 16 times the vertical. As was previously explained in section 114, if the square of the vertical is subtracted from the square of the hypotenuse the remainder is the square of the base. Accordingly, if the square of the vertical is subtracted here the remainder is 8 times 8 less 16 times the vertical, which equals the square of the base. If one adds onto this 16 times the vertical the remainder is 8 times 8 which is equal to the square of the base plus 16 times the vertical, as was written in section 46. So in this example, if the square of the base, which is 16, is subtracted from 8 times 8 the remainder is 48. If that is then divided by 16 times the vertical the result is that the vertical measures 3. The lesson learned here is that if one wishes to know the length of the vertical one need only square the connecting line (hypotenuse), subtract from it the square of the base, divide it by the hypotenuse and half of that result is the length of the vertical. This process can be shortened by omitting the squaring of the hypotenuse, and just dividing the square of the base by the hypotenuse, with the remainder then subtracted by the hypotenuse, as is known in algebraic math. Half of this remainder is then the length of the vertical, as is shown in this formula:

((a^2)/a)- (b^2) = ((b^2)/a)-a

##### Section 120Edit

All this applies to a right triangle. If the triangle is not a right triangle but an equilateral triangle there is only a need to know the measure of one side, and there is no need to know any of the angles.

##### Section 121Edit

If the triangle has two equal length legs (isosceles) there are only four quantities that need be known: The length of one of the legs, the length of the third side, one of the two equal angles, and the third angle. If two of these quantities are known the other two may be deduced. This is possible through five distinct arrangements: The length of one of the legs and the two equal angles are known, the length of one of the legs and the third angle are known, the third side and one of the two equal angles are known, the length of the third side and the third angle are known, or one of the two equal angles and the third angle are known. First though, one must draw a vertical line from the third angle to the base, with the two equal angles on the far sides being acute, as was written in section 100. The figure will therefore be as so:

as was shown in section 68.

##### Section 122Edit

Expanding on the above, according to the first arrangement mentioned, where the length of one of two equal sides (side AB) and the third side (side BD) are known, and a right triangle ACB is inscribed with sides AC, AB, and BC (which is half BD), the other quantities can then be determined, as was written in section 117. This goes for the triangle ACD as well, since lines AB and AD are equal, as are lines BC and CD.

##### Section 123Edit

In the second arrangement, where side AB and angle ABD are known, and angle ACB is a right angle, the other quantities in triangle ACB can then be determined, as was written in section 118. The same goes for triangle ACD as it is also equivalent to the other.

##### Section 124Edit

In the third arrangement, where the side AB and angle BAD are known, the angle BAC is half of angle BAD. Since this is so, the other quantities of these two triangles will be known, as is written in the previous section.

##### Section 125Edit

In the fourth arrangement, where side BD and angle ABD are known, side BC is half of side BD, and angles ABC and AGB are known, allowing the other quantities to be determined, as was written in section 118.

##### Section 126Edit

In the fifth arrangement, where the side BD and angle BAD are known, angle BAC is seen to be half of angle BAD, and side BC is half of side BD, and angle ACB become known. Since this is so, the other quantities can be known, as we have seen.

##### Section 127Edit

If the legs of the triangle are not equal (scalene) one needs to know three of the quantities, and they can be arranged as well five different ways: If two angles and the side between them are known, if two angle and one of the sides opposite them are known, if two sides and the angle between them are known, if two sides and one of the angles opposite them is known, or the lengths of all three sides are known. In all these arrangements one needs to determine the three quantities that remain unknown.

##### Section 128Edit

Method 1)

If two angles and the side between them are known, draw a right angle line from one of the known angles to the side opposite it, like so:

Here angles ABC, BAC and side AB are known. Right angle line AD is drawn. Now, since in the triangle ADB angles ABD, ADB and side AB are known, the other quantities can be determined. If so, angle BAD may be subtracted from angle BAC to determine angle DAC. Since angle ADC is known, the other quantities of triangle ADC can be determined. When the lengths of BD and DC are added together the length of BC is then known, and when angles BAD and CAD are summed the measure of angle BAC is known. This all works when the right angle falls within the triangle.

##### Section 129Edit

If the vertical line falls outside the triangle, like so:

where angles ABC, BAC and side AB are known, then in triangle ADB angles ABD, ADB and side AB are known. If so, the other quantities can also be known. Since these other quantities are known, line AD is also known and angle BAD is known as well. If angle BAC is subtracted from angle BAD then the remainder is angle CAD and ADC then becomes known as well. Once this point is reached the other quantities of triangle ADC can be determined.

If side CD is subtracted from side BD the remainder is side BC. Angle ACB is then the remainder of angle ACD and side AC can then be determined.

##### Section 130Edit

Method 2)

If two angles and one of the sides opposite them are known, draw a right angle line from the third, unknown angle to the side opposite it, like so,

where angle ABC, ACB and side AB are known, and vertical line AD is dawn, creating triangle ADB. Here the two angles ABD and ADB are known, as is side AB. Since this is so, the other quantities can be determined. Afterward, in triangle ADB angles ADC and ACD are known, as is side AD. If so, the other quantities can also be determined. Then the angles and sides are combined [back into the large triangle, its measurements are known,] as we have seen.

##### Section 131Edit

If the vertical falls outside the triangle, like so:

a line is also drawn from the angle that is not known [straight down]. For example, if angles ABC, ACB and side AB are known, and angles ADB, ABD and side AB of triangle ADB are also known, then the remaining quantities can be determined. After this, consider triangle ACD with method 1 where the vertical falls outside the triangle and side AC can be determined as well; though this approach is not necessary, as it has already been shown that the proportion of angle ACB to side AB is the same as the sine of angle ABC to side AC [see sections 87 and 88].

##### Section 132Edit

Method 3)

If two sides and the angle between them are known, draw a vertical from one of the two angles that are not known to side opposite like so:

Here sides AB, BC and angle ABC are known. The vertical line AD is drawn, creating triangle ADB. Angles ABD, ADB and side AB are now known. If so, the other quantities can now be determined. If line BD is removed from line BC the remaining line DC is known. Side AD and angle ADC of triangle ADC are known, and therefore the other quantities can be determined, as has been shown.

##### Section 133Edit

If the vertical falls outside the triangle, like so, where sides AB, BC and angle ABC are known:

In triangle ADB angles ADB, ABD and side AB are known. If so, the other quantities can be determined. If line BC is then removed from side BD what remains is line CD. At this point, in triangle ACD two sides are known, AD and CD, and one angle is known, ADC. If so, the other quantities can be determined, as we have shown.

##### Section 134Edit

Method 4)

If two sides and the angle opposite them are known: Draw a vertical line from the angle between the two known sides to the opposite side, like so:

Here sides AB and AC are known and angle ABC is known. Line AD is drawn forming triangle ADB. Angles ABD, ADB and side AB are known. If so, the rest can be determined. After this, the sides AD and AC and angle ADC in triangle ADC are known, and the remaining quantities can be determined, as we have shown.

##### Section 135Edit

If the vertical falls outside the triangle, like so, where sides AB and AC are known, and angle ABC is known:

In triangle ABD side AB and angle ABD and ADB are known. This means that the other quantities can be determined. After that, sides AC, AD and Angle ADC in triangle ACD are known, and that enables the remaining quantities to be determined, as we have seen.

##### Section 136Edit

Method 5)

If all three sides are known, draw a vertical line from one of the angles to the side opposite it, like so:

Here sides AB, AC and BC are known. The vertical AD is drawn and thereby the line BC is divided into two parts. The square of side AC is then added to the square of side BC. Then the square of side AB is subtracted from them. The remainder is divided by side BC and half of that amount is line DC. This can now be subtracted from side BC to yield line BD.

An alternate approach is to add the square of side AB to the square of side BC and then subtract from them the square of side AC. The remainder is divided by side BC and half of that amount is line BD. If that is subtracted from side BC the result is line DC. The proof of this is that the square of side AB less the square of side BD is equal to the square of the vertical line AD, as was written in section 75. Similarly, the square of side AC less the square of side DC is equal to the square of vertical AD. If so, both of these are equal to each other, as was written is section 47.

The square of side DC is equal to the sum of the square of side BC and BD less two times the product of side BC and BD, as was written in section 119. It is further shown in algebra that subtracting value A from equation D-C=B will result in equation C-B-D=A. If so, the square of side AC and twice the product of sides BC and BD, less the sum of the square of side BC and BD is equal to the square of side AB less the square of side BD. Add to either of these equivalent results the square of side BC and the square of side BD and the result is that the square of side AB plus the square of side BC is equal to the square of side AC plus two times the product of BC and BD. Through this process side BD can be determined, and the reverse would determine side DC.

##### Section 137Edit

If the vertical falls outside the triangle, like so:

Here the three sides AB, AC and BC are known and vertical line drawn is AD. [In this situation] side AB is squared and the sum of the squares of sides AC and BC are subtracted from it. The remainder is divided by the length of side BC and half of that result is line CD. If this is then added to line BC the result is the length of side BD. The proof of this is that the square of side AB less the square of side BD is equal to the square of side AC less the square of side CD, as was previously shown. In addition, the square of side BD is equal to the sum of the squares of sides BC and CD plus twice the product of side BC and CD, as was written in section 75. Since this is so, the square of side AB less the sum of the squares of sides BC and CD plus twice the product of sides BC and CD is equal to the square of side AC less the square of side CD. Adding on to either of these equivalent statements the sum of the squares of sides BC and CD plus twice the product of BC and CD will equal the square of side AB, which is also equal to the sum of the squares of sides AD and BC plus the product of sides BC and CD. After this has all been demonstrated it is clear that in the two triangles ABD and ACD, where three quantities are known, the other quantities can be determined regardless of whether the vertical falls inside or outside.

##### Section 138Edit

There is another approach that can be employed if the vertical falls within the triangle:

In this example, **[comment H]** add side AB and side AC, and subtract side AB (the smaller value) from the larger side AC. Then multiply the previous summation by the previous remainder and divide the result by side BC. Half of the result is the measure of the smaller portion of side BC, which is line BD. The reminder of BC less BD is then DC. Alternatively, one may add the result of the previous division to side BC and half of the result is equal to side DC. BC less DC is then BD. The proof of this that the square of side AB less the square of line BD is equal to the square of side AC less the square of line DC, as was shown in section 136. It is further explained via algebra that an equivalency B-C = B+A is equal to D+B = C+A, in that if D+B is added to the first equivalency the result is B+C = D+A. If C=D is then subtracted the result is D+B = C+A. accordingly, the square of AC less the square of AB is equal to the square of DC less the square of BD. In addition, it known through algebra **[Comment I]** that in a plane where one side's length is C+A and the other is C+A the area is (A x A) x (C x C). It can be shown through the following proof, in an example where side AB and AC have a length of A. and sides DB and CI have length of C. In the square ADIJ side AI has a length of C+A, and side AD has a length of C+A, and the square CHIJ is equal to square DBFE. If so, the sum of square AD CIH and square DBFE are equal to square ADIJ. If so, when square ABCD (which is equal to A x A) is subtracted from square FEHG (which is equal to C x C) the result is equal to square ADIJ. Since this is so, the square of AC less the square of AB is equal to the area of a plane where the length of one side is equal to AC plus AB and the second side's length is equal to AC less AB. Similarly, the square of DC **[MISPRINT OF 'DB' IN HEBREW]** less the square of BD is equal to a plane where the length of one of its sides is DC plus BD and the second side is DC less BD.

Here there exists a proportion: The two sides of the square plane made from AC less the square of AB may be placed at the extremes terms of a proportion and the two sides of the square plane made from DC less the square of BD may be placed in the middle terms and will be in proportion to each other, as was explained in section 44 [THAT IS, AC+AB:DC+BC = DC-BD:AC-AB, 'DC-BD' BEING THE UNKNOWN]. Now then, if the value of the square of DC less the square of BD is unknown one may multiply the two extreme terms of the proportion and divide the result by the one known middle term, as was written in section 45. Through this method the side DC less BD is known. If the result is subtracted from side BC the value is twice BD, while if the result is added to side BC the value is twice side DC. **[IS THIS CORRECT?]**

**[FROM HEBREW COMMENTARY:**

**[H]**For example, if side AC is 6, side BC is 5 and side AB is 4, 6 and 4 are summed to 10 and then 4 is subtracted from 6. This 2 that remains is multiplied by 10, equaling 20. 20 is then divided by 5, which equals 4. 4 is then subtracted from 5, equaling 1. Finally, half of 1 is the length of line BD, which is 1/2.

**[I]****[PARTIAL QUOTE]**...First I will describe what I believe to be the explanation of the text according to the author's intent and then I will add explanatory comments. This is his intent: If one is given an equivalency, such as (C-B=A) = (H-D=A) the components themselves must be equivalent. For example, C-B and H-D are equivalent, as was shown in section 47. Therefore, H-C is equal to D-B. The proof is that when one adds D-C on to both sides of the equation the result is that D-C+C-D is equivalent to D-C+H-D. The equivalency is not invalidated because the same term D-C is added to both sides, and positive C and negative C cancel each other out, for what is the difference whether C is added and then subtracted or whether both are crossed out? The positive and negative variable D also cancel each other out. At this point the remaining second term H-C is equal to the remaining first term of D-B. Accordingly, the square of side AC less the square of the side AB is equal to the square of line DC less the square of line BD. Until here is the intent of his words....

- ...On this, the statement "it is known in algebra" until the end of the section was stated, and since the language is so brief, I will set down the text in a way that truly reflects his intent [the author's], and then I will explain further:
- “It is known in algebra that if one side has the length C+A and second has the length C-A the plane formed is C x C - A x A. :It may also be shown in the following diagram:

- where sides AB and AC have the length A', sides DB and CI have the length C', rectangle ADIJ has a side AI which measures C'+A', and side AD is C'-A'. The rectangle CHIJ is equal to rectangle DBFE. If so, the rectangles AFCH and DBFE together are equal to rectangle ADIJ. If square FEHG, which measures C' x C', is removed from square ABCG, which measures A' x A', the result is equal to rectangle ADIJ. If so, the square of AC less the square of AB is equal to a plane where one of its side’s measures AC plus AB while the second side equals AC less AB. Similarly, the square of DC less the square of AB is equal to a plane one of whom's sides equals DC plus BD while the second side equals DC less BD. Now, if the two sides of the plane that equals the square of AC less the square of AB are placed at the extremes of a proportion and the two sides of the plane that equals the square of DC less the square of BD are placed as the middle terms they are in a correct proportion, as was written in section 44. Now if the square of DC less the square of BD is the unknown variable, the two extremes of the proportion should be multiplied and then the result should be divided by one of the known middle terms, as was written in section 45 [concerning proportions]. This method will produce the value of side DC less side BD. If this number is subtracted from side BC the result is two times side BD. If this number is instead added on to BC the result is two times DC.” Until here is the intent of his [the author's] words...]

##### Section 139Edit

If the vertical falls outside the triangle, like so:

where sides AB, AC and BC are known, with vertical AD drawn. Add side AB to side AC and subtract Side AC from side AB. Multiply the subtracted result by the summed result and divide that result by side BC. From the remainder subtract side BC and half of that remainder is side CD. The proof is this is found by squaring side AB and subtracting from it square of side BD. This is equal to the square of side AC less the square of side CD, as we have seen. Since this is so, the square of side AB less the square of side AC is equal to the square of side BD less the square of side CD. If so, there is a proportion here of AB+AC:BD+CD = BD-CD:AB-AC. If the two extremes are multiplied by each other and the product is divided by BC the result is the sum of BD and CD. If BC is subtracted from this number the result is two times CD. After this calculation all the three unknown quantities in the triangle become known, as has been shown previously.