# Translation:Ayil Meshulash/Discourse 5

**This chapter contains sections 140 to 162**

##### Section 140Edit

If one marks a point within a circle outside of center, like so:

where point A is the center and point B is the point outside the center, the greatest line that can be drawn from this point towards the circumference of the circle (point C) will pass through the center. This line is called the maximum **[WHAT IS THE CORRECT TERM?]**, for its end at the circumference is the farthest point in the circle from point B. If the rest of this line is completed as a diameter across the circle the point on the other end (point E) is called the minimum **[WHAT IS THE CORRECT TERM?]**, for it is the closest point in the circle to point B, and a line drawn from point B to point E is the shortest line that can be drawn from that point until the circumference. The proof is that any line drawn from point B until some other point, like point D, for example, must be shorter than line BC, as lines AD and AB together are longer than line BD, and line AD is equal to line AC. If so, line AD and line AB together are equal to line BC. Since line BD must be shorter than line BC, and line BD is longer than BE, since line BD and AB together are more than line AD, as we have shown, and Line AB and BE together are equal to line AD, line BD therefore must be greater than line BE, and all other lines extending from point B that travel away from the direction towards the upper or lower bounds are by force becoming smaller.

##### Section 141Edit

If line AB is divided into two equal parts, and a right angle line is extended until the circumference (at point F), this line is greater than half the diameter, all lines extending from point B to a point between the maximum and point F are greater than half the diameter, while all lines drawn from point F to the minimum (under point B) is less than half the diameter. A line drawn from point B to point F is half the diameter. Since line FG divides line AB in two parts at the right angle, the triangle ABF is an equal leg triangle (isosceles), as was written in Section 52, line AF is half the diameter, and therefore line BF is half the diameter as well.

##### Section 142Edit

Angle CAD is greater in all possible lines extending from the diameter [from point A] than angle CBD is in all the possible lines extending from point B when they both meet at the same point on the circumference, as was written in section 67. Similarly, angle DAE is smaller than angle DBE in all its possible paths, as was written there.

##### Section 143Edit

The difference [in degrees] between angle CAD and CBD is equal to angle ADB. The proof to this is to draw a line parallel line to AD from point B until the circumference, like so:

Here line BF is parallel to line AD, and angle CBF is equal to angle CAD, as was written in section 20. If the angle CBD is subtracted from these angles the remainder is angle DBF, and DBF is equal to angle ADB, as was shown in section 62.

##### Section 144Edit

This can be proven through an alternate method as well. As angle CAD is equal to the sum of angles ABD and ADB, as was written in section 65. This being so, angle CAD is greater than angle ABD in the amount of angle ADB, and the same result will always result when lines are drawn from along the diameter line from the maximum to the minimum, and this will be the difference between angles DAE and DBE in every arrangement [as angle DBE contains the sum of the degrees of angles DAB and ADB].

##### Section 145Edit

If the distance between the center and the point along the diameter is known (line AB in above figure), it is possible to determine the angle in degrees of angle ADB for every degree from the maximum to the minimum [top of diameter to the bottom]. In addition, it is possible to determine length of line BD wherever it may be, provided it extends from the point to the surface of the circle. This is due to the fact that a) Angle CAD is known, b) Angle DAB is known, as it is the remainder [to 180 degrees], as was written in section 15, c) Line AD half the diameter, and d) Line AB is known. With these values known the other quantities can be determined, as was written in section 133. Similarly, if the opposite angle [ADB] is known, the distance between the center and the point can be determined.

##### Section 146Edit

The opposite angle [AIB] grows larger as it sweeps from the maximum to the minimum. At point I, the line extending to point B on diameter line CD forms a right triangle, and as the angle sweeps past point I it becomes increasingly smaller. Therefore, point I is point where the sweeping angle's measure is the greatest.

The proof of this is that the proportion of line AI to line AB is equal to the proportion of the sine of angle ABI to the sine of angle AIB [text corrected per commentary]. This, and the fact that lines AI and AB are always the same length during the sweep cause the sine of angle AIB to enlarge if the sine of angle ABI is enlarged. Since the measurement of the sine of a right angle is the greatest of all angles, as was written in section 83, and this is angle ABI at point I, angle AIB and point I is the greatest of all opposite angles along the sweep.

Additionally, the measure of the sine of angle AIB at point I [which is line AB, to the sine of angle ABI, line BI] is equivalent to the proportion of line AB to half the diameter, as was written in section 87. **[IS THIS CORRECT?]**

The situation is reversed when sweeping from the minimum to the maximum, where from point D to I the opposite angle increases and from point I to the maximum it decreases.

##### Section 147Edit

If one extends lines from the center of the circle to every possible angle along the circumference of the circle, and compares them to lines extended from a point removed from the center to every degree [and compares the two sets of lines,] one will find that from the maximum to point I [in previous section] the angle created by the center point always exceeds the angle created by the other point, while from point I to the minimum the angle created by the other point always exceeds the angle creates by the center point. For example, angle EAF is greater than angle EBF, and angle GAH is less than angle GBH.

##### Section 148Edit

The closer an arc is to the maximum the greater will be the difference between two angles, such as EAF and EBF [created by angle formed from the each end of the arc to the center and a point outside the center]. As the arc sweeps downward the difference between the two angles decreases until point I. Similarly, the closer an arc is the minimum the greater the difference between two angles, for example GAH and GBH, and that diminishes as the arc continues to sweep to point I.

##### Section 149Edit

Everything we have mentioned regarding the area to the left of the diameter CAD is true for the right of the diameter as well. Therefore, if you calculate 180 degrees of angle and arc from the maximum you will reach the minimum, and from 180 degrees to 360 degrees you will pass from the minimum to the maximum.

##### Section 150Edit

If you draw a small circle that passes through the circumference of the large circle and then draw a line from the center of the large circle across the small circle to its circumference, the longest line that can be drawn is on that passes through the small circle’s center, like so:

Here, where circle ABCD is drawn over the circumference of circle ADK, the longest line that can extend from the center at point H until the circumference of the small circle is line HB. Point B is called the maximum. The shortest line that can be drawn to the small circle from the center is line GH, and point G is called the minimum. The proof to this rests on what was mentioned in section 140, that if a line such as HC is drawn it is perforce smaller than line HB, since lines HF and HE are equal, and lines BF and CF are equal, and the sum of lines HF and CF is greater than line HC. Since this is so, line HB is also greater than line HC, and line HC is greater than line HG. The proof to this is similarly found by referring to that section, for the sum of lines HC and FC is greater than line HF and greater than the sum of lines HG and CF, which is equal to line HF. This follows the idea that all lines drawn out [as one sweeps around the circumference] from the maximum to the minimum decrease in length. All the rules explained from section 140 to section 149 can be found in this example, as will be explained.

##### Section 151Edit

[In the figure above,] all the lines that extend from the maximum until the circumference of the large circle between points D and A are all larger than half of the diameter of the large circle, which is line HF in the figure above. From between these points [along the circumference] until the minimum all lines will drawn to the minimum be shorter than half of the diameter, and at these two points lines drawn would be equal to half the diameter, as was written in section 21. **[THE COMMENTARY IS UNABLE TO RELATE THIS SECTION TO SECTION 21, AND SUGGESTS THAT SECTION 7 MAY BE THE CORRECT TEXT.]**

##### Section 152Edit

In the figure above, the difference between angle BFC and angle CHF is angle HCF. The proof to this is also follows from what was mentioned in section 143, and the same is true for every point along the circumference of the small circle, where the angle continually grows [as the point moves] from the maximum to the minimum. Angle HFC then has the opposite property, for it decreases as it sweeps from the maximum toward the minimum. At the same time, angle CHF increases as it sweeps from the maximum to the point where a line extending from the center of the small circle forms a right angle with a line that extends from the center of the large circle. This occurs at point I. The proof of this also follows section 146, for the proportion of half the diameter of the large circle to half the diameter of the small circle is equivalent to the proportion of the sine of angle FCH to the sine of angle CHF. Therefore, since the halves of the diameters of each circle always measure the same lengths respectively, the size of angle CHF is wholly dependent on the change of the sine of angle FCH. Now the greatest of all sines is that of the right angle [which angle HCF forms], and if so, the angle CHF must be the largest of all the angles. This means that from that point on they decrease until the minimum is reached, while from the minimum to the maximum the opposite is true.

##### Section 153Edit

If one knows the proportion of the diameter of the small circle in relation to the diameter of the large circle, and also knows how many degrees away a point is [along the circumference] from the maximum, which [in our example] is angle BFC, one is also able to determine all three angles of the triangle CHF, and the length of side FH, since angle CFH is known to be the remainder of angle BFC. In addition, the lengths of sides CF and HF are then known. Since this is so, the other quantities can be determined as well. Similarly, if one knows the proportion of the diameters and angle CHF [and angle BFC is unknown], the remaining quantities can be determined. Furthermore, if angle BFC and CHF are known and the diameter of the large circle is also known the other quantities can be determined, and the same is true if the other quantities are known.

##### Section 154Edit

It is also clear from what was explained in section 152 that regarding angle CHF, when it is at its greatest point [and angle HCF is a right angle, so that line FC reaches point I,] the proportion of half of the diameter of the large circle [line HF] to half of the diameter of the small circle [line CF] is equivalent to the proportion of the sine of angle HCF to the sine of angle CHF.

All the principles explained [to this point] regarding a point lying distant from the center relate to the figure outlined in section 150.

##### Section 155Edit

We have previously explained that through every unit of change from the maximum until point I the angle CHF increases. However, the increase in degrees per unit change is not constant. The closer to the maximum the greater the angle measure. The same is true for the minimum. In this example,

the arc points are B, K, and C. When drawn further out as another arc with the center at H and the radius extending to B, arcs BK and KC are equal in arc measure and lines drawn out from point H would create an equal pair of angles. However, these same changes of degree on the arc of the small circle would create pair of angles that are do not equal each other when the lines are drawn out from point H. When lines are drawn out to these points on the small circle from point H, angle BHK is greater than angle KHC, [and the rate of change in the increase of angle CHF is diminishing as it sweeps left]. The same is true for the minimum, that though arcs GL and LM are equal, angle GHL is greater than angle LHK. All that has been written in regard to the small circle is true for the right side of this circle as well. Therefore, if 180 degrees are calculated from the maximum the above can all be calculated from the maximum to the minimum, and if 180 degrees are added from that point until 360 degrees the same calculations can be made from the minimum to the maximum. **[THE TRANSLATION OF THIS SECTION IS UNCLEAR. IT HAS BEEN RENDERED IN A WAY THAT SEEMS SENSIBLE, BUT IT MAY OR MAY NOT BE AN ACCURATE TRANSLATION OR CORRECT.]**

##### Section 156Edit

If the [circumference of the] circle lay outside the center of the circle, similar to [the idea depicted in] section 140 **[UNSURE OF THIS PHRASE]**, and it is equal to the large circle in the previous section, and the difference between the center and a point outside the center is equal to half the diameter of the small circle: Since this circle is equal to the circle in the previous section and line TS here is equal to half the diameter, which is line BF, all the six quantities of triangle TRS are equal to the six quantities of triangle CFH in every degree as R becomes distanced from N, and C becomes distanced from B, respectively, if the degree changes are equal. Angles CFH and RTS are equal, for they are the remainder of arcs BC and NR. Lines FH and TR are also equal [both are radius lines], and lines CF and TS are equal as well. Therefore, all the other quantities [of these two triangles] are equal, as was written in section 24.

##### Section 157Edit

If angle VHM is equal to angle EMP and line HY is equal to half the diameter of the small circle which is line MP, then it follows that line YP is equal to line HM. The proof of this is that since angles VHM and EMP are equal the result must be that lines YH and PM are parallel, as was written in section 20. Consequently, angles PHY and HPM are equal, as was written in section 62, and lines HY and PM are equal, with line PH being shared by the two triangles PYH and PMH. If this is so, all the quantities of these triangles are equal, as was written in section 24.

It further follows that lines YP and HM are parallel since angles YPH and PHM are equal, as was written in section 62. In addition, it is evident that the angles VYP and VHM are equal, since the lines YP and HM are parallel. All of this is true in regards to all the parts of the large circle in which is set the center of a small circle, for point M in relation to point V in the large circle is as the relation of point P to point E in the small circle, and as a result the two arcs VM and PE are equal.

##### Section 158Edit

From here it is apparent that if a small circle is inscribed so as to overlap a large circle, and point P is placed on the circumference of the small circle along the same vertical line as the center of the small circle, and then this point P is used to draw a another circle, such that the new circle is the same size as circle ABKCZ and its center is at point Y (as it has been explained in the previous section that line YP is equal to line HM, which is half the diameter), then the following is so: Point Y is the center, point H is outside the center, the triangle YPH in the new circle is completely identical to triangle TRS [in section 156] in all its particulars within circle NRE **[Where is the ‘Hey’ in the diagram?]**. This is because lines YH and TS are equal, lines YP and TR are equal, and angles PYH and RTS are equal. These angles are also the respective remainders of angles VYP and NTR. Therefore, all the respective quantities of these triangles are equivalent. The same is true for every part of these circles, so that the new circle can be placed on circle NRE **[Where is the ‘Hey’ in the diagram?]** in an orientation that would allow triangle YPH to be exactly positioned over triangle TRS. The same maximum and minimum points [along the diameter] that connect the lines for TRS would also connect the lines for YPH and are therefore equal, with lines PY and PH corresponding to lines RT and RS, providing that the degrees measure from point M to point V on the left, and point R from point N on left, and point P from point E on the right are equal, or alternatively, the degree measure between point M and points R (on the right) and P (on the left) are the same.
**[NOT SURE WHAT THE REFERENCE TO ‘RIGHT’ AND ‘LEFT’ MEAN]**

##### Section 159Edit

All of the above applies to lines extending from the center. If, however, the lines extend from a point outside the center, arcs that are of equal measure in two different locations on a small circle will not create equal angles. In this figure,

lines TY, BY, HY and VY all extend from point Y, which is removed from center P. While arcs AB and HV are equal, angle HYV is nevertheless greater than angle AYB. Similarly, as the distance from the maximum increases so does the angle measure, up until minimum point M where the angle is at its largest measure. From the minimum until the maximum the angle decreases until it reaches the maximum, where it is at its smallest measure. **[NOT SURE OF THIS SECTION]**

##### Section 160Edit

To determine this initial angle measure (of angle HYV), one must know the measure of the line that extends from point Y until the center of the small circle, which is line CY. This is known by use of the triangle PYC, as was explained in section 145. When line HC and angle HYC are known the remaining quantities of triangle HCY [per commentary] can be determined.

[Commentary: This means to say that after one has determined angle APC, which presupposes that the angle measure between center C of the small circle and point T of the large circle is known, then one can determine angle CPY, which is the unknown angle’s (APC’s) remainder. Now line PC is known to be half the diameter, and line PY can also be known through the application of section 145. If all this is so, line CY can be determined and line HC is also known, as it is half of the diameter of the small circle. Angle HCY can also be known, as we have seen it is the remainder of angle HCV, for we know the angle measure between point H and V in the small circle. Therefore, the angle HYV [the angle this section is interested in] can be determined.]

##### Section 161Edit

An alternative method that can be used, providing lines PY and HC are equal: Set the following proportion: The sine of angle CPY compared to the sine of angle PCY is proportional to the sine of angle CHY to the sine of angle CYH. Now angles CPY and PCY can be known, as was written in section 145. Once the proportion of angles CHY and CYH are known, and the summation of angles CHY and CYH are known to be equal to angle HCV (as was written in section 65), then the measure of each of the angles CHY and CYH can be determined (as was shown in section 95) [and angle CYH is equal to angle HYV].

The proof of this is that the sines are proportional to each other, for as is the proportion of line CY to line PY so is the proportion of the sine of angle CPY to the sine of angle PCY (as was shown in section 87). Furthermore, the proportion of line CY to line HC is as the proportion of the sine of angle CHY to the sine of angle CYH. The line PY of this proportion is then equal to the line HC of the other proportion. Therefore, proportion of these pairs of angles must be equal to each other [CPY:PCY = CHY:CYH].

Through this approach any arc on the small circle can be used to determine the degree measure of any arc on the large circle from the maximum to the minimum and from the minimum to the maximum [through 360 degrees].

##### Section 162Edit

If two lines extend, one from the center of the large circle and one from a point outside the center, and both pass through the center of the small circle until its circumference, like so,

where lines BH and CH extend respectively from center B and point C to center H: In this formation one can determine arc GI at any degree of the large circle providing on knows the measure of arc AH. In addition, the following can be known: line BC, line BH (which is half of the diameter), angle BHC, and that angles BHC and GHI are equal. Know further that all I have written in regard to a point that is outside the center from section 140 to 150 will apply to angle GHI here since angles BHC and GHI are equal and triangle BHC is equal to triangle ADC from section 140. The same is true for angle FHI that comes about through lines BH and DH and for the two lines extending from points outside the center, lines CH and DH, which create angle FHG. However, in this later instance either triangle CHD must be calculated first before triangle BHC is calculated so that line HC can be known, as was written in section 160, or, if lines BC and CD are equal, the proportion of the sines of the angles method may be used, as was written in the previous section.