Translation:Ayil Meshulash/Discourse 3

Ayil Meshulash , translated from Hebrew by Wikisource
Discourse 3
Geometric Theorems

This chapter contains sections 48 to 80

Section 48Edit

If you wish to assign symbols to a shape to denote a triangle, place the letters at each corner. Then, if one wishes to speak of a specific angle he will place the angle in question the middle of the triangle’s letter symbol [In triangle ABC, angle C is called angle BCA]. If one wishes to speak of one of the edges, they are called by symbol of the two angles at each end [side BC or AB, for example].

Section 49Edit

The area of a plane is how many square cubits, square handbreadths, or square finger-lengths are contained inside, whether the plane is square, circle, triangle, or any other shape imaginable.

Section 50Edit

The area of a square is well known, as it is the multiplication of the vertical edge by the horizontal edge.

Section 51Edit

In an ‘even footed’ (isosceles) triangle, the angles opposite the equal edges are themselves equal. For example, in triangle ABC, side AB is equal to AC, and angle ABE is equal to ACE. The proof to this is that if isosceles triangle BCD is constructed on the opposite side of line BC, and line AD is shared as a side of both triangles ABD and ACD, angles BAD and CAD will be equal to each other, as demonstrated in section 23. Following this, the two triangles ABE and ACE are equal, since the two sides AB and AC are equal, and angles BAE and CAE are equal, with line AE shared between the two. Therefore, angle ABE and ACE are equal, as was written in section 24.

Section 52Edit

From here it is also clear that line BE is equal to line CE.

Section 53Edit

It is also clear that angle BEA is equal to CEA, and if so, they are both right angles, for both are 180 degrees, and was explained in section 15.

Section 54Edit

From this it is further known that if one wishes to divide a line segment into two equal segments by the use of a right angle one should draw two isosceles triangles on both sides of the line, then draw a line from each vertex angle to the line being divided [the base that they both share], as was explained in section 51 where line BC is bisected by line AD. Alternatively, one may draw arcs from either end of the base on both sides (as was explained in section 28), and a line drawn bisecting the base from the points where the arcs intersect.

Section 55Edit

If one wishes to make a line that is parallel to an existing line, cut the first line by another at a right angle and then cut that intersecting line by a third line at a right angle at some distance away. The resulting third line is parallel to the first, like so:

Section 56Edit

If one wishes to ascertain the center of any circle one should draw a chord across the circle from any point and then bisect that line with another at a right angle. The line that bisects is perforce the diameter of the circle. Finally, bisect that line with another. The point where these two lines cross is the center.

Section 57Edit

It was explained in section 51 that when all of a triangle’s sides are the same length all of its angles are identical as well, as seen in this figure:

Section 58Edit

If two of a triangles angle are the same then the sides facing those to angles are the same length as well, as seen in a previous figure, where angle ABE and ACE are equal, and side AB and AC are equal as well. This can be proven by constructing an isosceles triangle on the opposite of the base of the first triangle, so that angles EBD and ECD are equal, as was seen in section 51. Therefore, angles ABD and ACD are equal as well as sides BD and CD. Side AD is shared between the two triangles ABD and ACD, and sides AB and AC are equal as well, as was seen in section 24.

Section 59Edit

From this it is clear that a triangle whose angles are all the equal to each other also has sides that are all equal to each other.

Section 60Edit

Furthermore, it is possible to prove this with a simpler proof, by drawing the line segment AE in triangle ABC, thereby creating right triangles AEB and AEC, which are equal, as was written in section 15. Side AE is now shared by the two triangles ABE and ACE, which are equal to each other, and sides AB [??] and AE, and angles ABE and ACE, are equal respectively. Therefore, the remaining [angle and side] is equal as well, and was written in sections 24 and 25.

Section 61Edit

Two lines that intersect create vertical angles that are opposite each other and are equal to each other. In the example below angles AEB and CED are equal. Since angles BED and AEC are their remainder they are also equal to each other, as was written in section 46.

Section 62Edit

A line that transverses two parallel lines creates corresponding angles that are equal to each other, like so (as was written in the previous section):

Here angles ACD and CFG are equal. Therefore, whenever the angles are equal [NOT SURE ABOUT THIS SECTION] the lines are parallel. Also, angle DCF [text missing, probably ‘and GFH are equal’].

Section 63Edit

Together, the three angles of a triangle comprise 180 degrees exactly. The proof of this is found when a line drawn at one corner that is parallel to the side opposite it, like so:

Angle ABD is equal to angle BDE, as was seen in the previous section. Angle CBE is also equal to angle BED. All these angles (ABD, CBE, and DBE) together contain 180 degrees. Therefore, the triangle contains 180 degrees.

Section 64Edit

It is understood from this that there can never be a triangle with a right angle and an obtuse angle, nor one with two obtuse angles.

Section 65Edit

It is further understood that if one extends the line of one of the sides of the triangle the outside angle will contain the number of degrees present in the other two angles, like so:

Angle CBD is equal to the sum of angles CAD and ADB, since angle ABD is the remainder of angles CAD and ADB as well as angle CBD [in terms of reaching 180 degrees]. Therefore they are equal to angle CBD, as was written in section 15.

Section 66Edit

All the angles of a square, summed together, equal 360 degrees. If one draws a line diagonally from one corner of the square to the other corner the square has been divided into a triangle. If so, it is shown to contain 180 degrees twice. A pentagon contains 180 degrees three times, for it can be divided into three triangles. A hexagon contains 180 degrees four times. The general rule is that if one subtracts two sides from the total number of sides of any shape the result is the number of times in contains 180 degrees.

Section 67Edit

Two lines that originate from a point and meet a straight line: The longer line has the smaller angle than the shorter line, and the opposite is true on the complementary angle [IS THIS THE RIGHT TERM?], like so:

Angle ACE is smaller than angle ADE, and angle ACB is greater than angle ADB.

Section 68Edit

It is understood from here that if a vertical line is drawn from one corner of a triangle to the side opposite it, and the two other angles are acute, the line will fall within the triangle, like so:

Section 69Edit

If one of the other angles is obtuse, the vertical line will fall outside the triangle, like so:

Section 70Edit

If one of the angles is a right angle, the line that is next to the right angle is the vertical line itself, like so:

Section 71Edit

The area of a triangle is half the area of a square [UNSURE OF NEXT FEW WORDS]. The proof of this is to draw a line from the opposite vertical leg to the base, like so:

lines AB and CD are equal, lines AC and BD are equal, and line AD is shared by both triangles ABD and ACD. Since this is so, the triangle ACD is half of the square ABCD.

Section 72Edit

If the vertical leg is found to be inside the triangle like so:

then construct a line parallel to the vertical and line parallel to the horizontal, as in triangle BDE. This triangle is half of square BCEF. If so, triangle BDF is half of square ACDF.

Section 73Edit

If the vertical leg falls outside the triangle [CDE], and one extends the horizontal floor until the [virtual] right angle leg is met [as far as the triangle extends, in the example below on the far left], and then constructs lines parallel to the vertical and horizontal, as in this example:

Now if one removes triangle ACD, which is half the square ACDF, and triangle CEF, which is half of the square BCFE. What remains is triangle CDE, which is half of square ADBE.

Section 74Edit

Another proof [in considering triangle BFE] would be to extend lines vertically and horizontally from the two smaller angles of the triangle towards each other until they meet, and then continue to extend the horizontal line that is parallel to the base for another length as long as the base line. After this a line should be drawn from the far end back to the wide angle of the triangle. The figure would be like so:

In this figure the following is true:

1) Lines AB and CD are equal

2) Line BC is shared by both lines above and therefore lines AC and BD are equal.

3) Lines CF and DE are equal.

4) Angles ACF and BDE are equal.

Now if triangle BCG is removed two four sided shapes remain: AFBE and CFDE. These two shapes are equal. In addition:

1) Lines AF and BE are equal.

2) Lines AB and FE are equal.

3) Line BF is shared by triangles ABF and BFE.

Due to the above, triangle BFE is half of square AFBE and half of square BFDE.

Section 75Edit

In considering a right triangle, the area contained within a square constructed from the length of its hypotenuse is equal the area of the squares constructed from the lengths of its vertical leg and base respectively. The proof of this is found when a square is created by doubling the vertical leg and the base. Triangles ABD and BDE are equal, and triangle BDE is half of the square ABDE. The same is true for triangles BEF, FEH, and DEH. As a consequence of this, all the triangles in the square BDHF equal twice the area of square ABDE. This is the method if the vertical leg and the base are the same length. If they are not, the proof is then to construct a square from surrounded by these triangles, using the base of one and vertical leg of the other as the length of each side, resulting in an inscribed triangle constructed of the four hypotenuses, like so:

Lines AB, CE, DF, GH are the long sides, lines BC, ED, FG, HA are the short sides, and lines HE, EB, BH, and FE are the hypotenuses. Now it is explained in the science of [WHAT IS THIS WORD?] that if one adds two of the sides together and squares the sum the result is the same as if one squares each of the two sides separately, sums them together and then adds twice the the result of one side multiplied by the other [(A+B) ^2 = (A^2+B^2) +2(A*B)]. This can also be proven geometrically. The following square ACFI is constructed by squaring two sides individually, resulting in squares BCFE and DEGH:

Now these same two sides are multiplied by each other and result is doubled, resulting in the squares ABDE and EFHI. Take these results and refer back to the previous figure, removing from the square the four triangles ABH, BCE, DFE, and GHF, each of which contains half of the area obtained by multiplying the two previously mentioned sides together. The resulting square BHFE is equal to sum of each of the two sides’ squares [the original contention being that the area of a square constructed from the hypotenuses is equal to the sum of the area of two squares constructed from the vertical and base respectively].


The the text of the proof here appears to be in disarray, therefore I have decided to show the proper reading of this proof, as is found in the work of Euclid in lesson 35. A more appropriate reading of the text should therefore be: 'Another proof would be to continue a line from the small angles' ( meaning angles EBF and BEF of the triangle BFE) 'which meet as a right angle,' (meaning lines BD and ED) 'along with a line parallel to the base of this triangle that is the length of the base.' (meaning line BA is the length of line EF) 'Then a line should be drawn from the far end' (from point A) 'until the wide angle.' (which is angle BFE) 'Then line FC should be drawn from the end of line FE so that it is parallel to line ED. It is then found that lines AC and BD are equal' (since lines AB and CD are both equal to line FE they are also equal to each other, as per rule 47. Line BC is then the bridge between them.) 'and lines CF and DE are equal.' (according to rule 22) ‘If triangle BCG is removed from the two triangles' (that is, triangles ACF and BDE) 'two four sided shapes remain, ABGF and CGDE, which are both equal. If triangle GFE is added to them AFBE and CFDE are shown to be equal. Additionally, lines AF and BE are equal and lines AB and FE are equal, with line BF shared between the two triangles ABF and BFE. This being so, triangle BFE is half of the four sided shapes AFBE and CFDE.']
Section 76Edit

This can also be proven via an alternate method: Construct squares out of the two sides of the triangle [after the square is divided into two triangles] and a square out of the hypotenuse. Then draw a line from the vertex where the vertical and the base meet through the hypotenuse to create a right angle with the hypotenuse and to divide the hypotenuse’s square into two rectangles, with the length of the hypotenuse’s side of one of the rectangles being equal to the length of the long side of the triangle and the length of the of the hypotenuse's side of the other rectangle being equal to the short side of the triangle, like so:

At this point, square ABCD is the square made from side CD, which is the large side of the triangle, and square DEHI is made from side DH, which is the short side of the triangle. Square CHJL is made from side CH, which is the hypotenuse, and line DK is the line from the vertex that meets line JL forming a right angle.

Now it can be concluded that the area of rectangle CFJK is equal to the area of square ABCD, and the area of rectangle FHKL is equal to the area of square DEHI. The proof of the first clause is that when one draws a the lines AH, CI, DJ, and DL it is found that line CH and CJ are equal and angles HCJ and ACD are equal, since they are both right angles. If angle DCH is added to both angles it is found that ACH and DCJ are equal (as was shown in section 46). This being so, the two triangles ACH and DCJ are equal, as was shown in section 24, and the triangle ACH is half of square ABCD, as was shown in the previous sections (73 and 74). Triangle CDJ also contains half the area of rectangle CFJK. Therefore, the square ABCD and the rectangle CFJK are equal in area.

The proof of the second clause follows a similar pattern: Lines DH and HI are equal, lines CH and HL are equal, and angles CHL and DHI are equal, both being right angles. If angle CHD is added to both it is found that angles CHI and DHL are equal. If this be so, the two triangles CHI and DHL are equal, and the triangle DHL has half the area of rectangle FHKL. If so, the square DEHI and the rectangle FHKL have the same area.

Therefore, the original contention is proven, that square CHJL has same area as the sum of the areas of squares ABCD and DEHI.

Section 77Edit

The circumference of a circle is more than three times the size of its diameter. The proof of this is found by dividing the diameter into four equal parts, drawing two chords across the two quarters points of the diameter as right angles and then drawing six chords from six points, each originating from one of the points created along the circumference of the circle by the diameter and the two chords, connecting each point to its neighbor. Then one should draw a line from the center of the circle to one of the points along the circumference created by the two original chords. The figure should appear as so:

Here the diameter is divided into four quarters via points D, C, and I. Lines BDE and FIH cross the diameter at right angles, allows six chords to be created at the end points: BA, AE, EH, HG, GF, FB. After this, line BC was created. The result is that lines AD and CD are equal and angles ADB and CDB are equal, since they are both right angles. Line BD is then shared by the two triangles ADB and BDC. Lines AB and BC are also equal, and BC is half of the diameter, as was written in section 7. This being so, line AB is also half of the diameter, and line BF is parallel to line DI. If so, line BF is also half of the diameter and the same is true for lines FG, GH, HE, and EA. If so, these six chords equal three times the diameter, and every arc is always longer than a chord drawn below it. Therefore, it is proven that the circumference is greater than three times the diameter. The actual measurement is said to be approximately 22/7. The amount that the diameter is smaller than the circumference is then approximately 7/22.

Section 78Edit

Another proof that can be employed here is to see that chord AB is half of the diameter, angle ACB is 60 degrees, angles ABC and BAC contain the remaining 120 degrees and lines AC and BC are equal. This being so, angles ABC and BAC are also equal, as was shown in section 51. Therefore, each angle is 60 degrees and this triangle BAG is an equilateral triangle, as was shown in section 59. The result is that length of all the sides together is equal to the diameter, and the same would be true for all six chords.

Section 79Edit

The area of a circle is equal to the amount gained by A) multiplying half of the diameter by half the circumference or B) multiplying a quarter of the diameter by the whole circumference or C) multiplying the whole diameter by a quarter of the circumference. The proof of this is cut the circle from the center until the circumference, like so:

In this way the line AB represents the length of the circumference and the vertical line DC is half the length of the diameter. It has already been shown in section 71 that the area of a triangle is half of the base multiplied by the height.

Section 80Edit

From this it is clear that the ratio of the area of a square [in which a circle is inscribed] to a triangle [created from the circle’s area, as in the previous section] is 14:11, and the ratio of this triangle to this square 11:14. For example, if the diameter is 14 cubits in length the circumference is 44 cubits, as was shown in the previous section. Now, a quarter of the circumference multiplied by the whole diameter is then shown to be 11 times the 14 cubits, while a square would be 14 times 14.