The Book of Lemmas: Proposition 12
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A Mathematical Droodle
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Copyright © 19962017 Alexander BogomolnyThe applet attempts to suggest Proposition 12 from Archimedes' Book of Lemmas:
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Proof
Let TR produced meet AB in M, and join PA, QB.
Since the angle APB is right.

Add to each side the angle RBQ, and
∠PAB + ∠QBA = (exterior)∠PRQ. 
But
∠TPR = ∠PAB, and ∠TQR = ∠QBA, 
in the alternate segments; therefore
∠TPR + ∠TQR = ∠PRQ. 
It follows from this that
TP = TQ = TR. 
[For, if PT be produced to O so that TO = TQ, we have
∠TOQ = ∠TQO. 
And, by hypothesis,
∠PRQ = ∠TPR + ∠TQR. 
By addition
∠POQ + ∠PRQ = ∠TPR + ∠OQR. 
It follows that, in the quadrilateral OPRQ, the opposite angles are together equal to two right angles. Therefore a circle will go round OPQR, and T is its centre, because
Thus
∠TRP = ∠TPR = ∠PAM. 
Adding to each the angle PRM,

Therefore
∠APR + ∠AMR = (two right angles), 
whence
∠AMR = (a right angle). 
Remark
Nathan Bowler comments from a distance of more than 2200 years:
Let M be the base of the perpendicular from R to AB, and let M' be the point of intersection of TR with the circumcircle of TPQ. Then ∠RM'P = ∠TM'P = ∠TQP = ∠QAP = ∠RAP so RM'AP is cyclic, as is RMAP since both ∠RMA and ∠RPA are right angles. Similarly, both RM'BQ and RMBQ are cyclic, so each of M and M' lies at the other point of intersection of the circles RAP and RBQ to R. That is, M = M', which gives the desired result as well as the extra fact that MPTQ is cyclic. Alternatively, it is suggestive to add a point C at the intersection of AP and BQ. Let O be the midpoint of AB, which is also the centre of S, and note that since TPO and TQO are both right angles T is diametrically opposite O on the ninepoint circle of ABC. That is, T is the Euler point corresponding to C. R is the orthocentre of ABC, so TR is the altitude from C, and is therefore perpendicular to AB. Whilst relying on the ninepoint circle, this proof illuminates what is going on in the figure. Indeed, it was this proof which suggested the construction of M' used in the previous proof. 
Archimedes' Book of Lemmas
Proposition 1: If two circles touch at A, and if CD, EF be parallel diameters in them, ADF is a straight line.
Proposition 2: Let AB be the diameter of a semicircle, and let the tangents to it at B and at any other point D on it meet in T. If now DE be drawn perpendicular to AB, and if AT, DE meet in F, then
DF = FE. .Proposition 3: Let P be any point on a segment of a circle whose base is AB, and let PN be perpendicular to AB. Take D on AB so that
AN = ND. If now PQ be an arc equal to the arc PA, and BQ be joined, then BQ, BD shall be equal.Proposition 4: If AB be the diameter of a semicircle and N any point on AB, and if semicircles be described within the first semicircle and having AN, BN as diameters respectively, the figure included between the circumferences of the three semicircles is "what Archimedes called arbelos"; and its area is equal to the circle on PN as diameter, where PN is perpendicular to AB and meets the original semicircle in P.
Proposition 5: Let AB be the diameter of a semicircle, C any point on AB, and CD perpendicular to it, and let semicircles be described within the first semicircle and having AC, CB as diameters. Then if two circles be drawn touching CD on different sides and each touching two of the semicircles, the circles so drawn will be equal.
Proposition 6: Let AB, the diameter of a semicircle, be divided at C so that AC = 3/2·CB [or in any ratio]. Describe semicircles within the first semicircle and on AC, CB as diameters, and suppose a circle drawn touching the all three semicircles. If GH be the diameter of this circle, to find relation between GH and AB.
Proposition 7: If circles are circumscribed about and inscribed in a square, the circumscribed circle is double of the inscribed square.
Proposition 8: If AB be any chord of a circle whose centre is O, and if AB be produced to C so that BC is equal to the radius; if further CO meets the circle in D and be produced to meet the circle the second time in E, the arc AE will be equal to three times the arc BD.
Proposition 9: If in a circle two chords AB, CD which do not pass through the centre intersect at right angles, then
(arc AD) + (arc CB) = (arc AC) + (arc DB) .Proposition 10: Suppose that TA, TB are two tangents to a circle, while TC cuts it. Let BD be the chord through B parallel to TC, and let AD meet TC in E. Then, if EH be drawn perpendicular to BD, it will bisect it in H.
Proposition 11: If two chords AB, CD in a circle intersect at right angles in a point O, not being the centre, then
AO^{2} + BO^{2} + CO^{2} + DO^{2} = (diameter)^{2} .Proposition 12: If AB be the diameter of a semicircle, and TP, TQ the tangents to it from any point T, and if AQ, BP be joined meeting in R, then TR is perpendicular to AB.
Proposition 13: If a diameter AB of a circle meet any chord CD, not a diameter, in E, and if AM, BN be drawn perpendicular to CD, then
CN = DM. Proposition 14: Let ACB be a semicircle on AB as diameter, and let AD, BE be equal lengths measured along AB from A, B respectively. On AD, BE as diameters describe semicircles on the side towards C, and on DE as diameter a semicircle on the opposite side. Let the perpendicular to AB through O, the centre of the first semicircle, meet the opposite semicircles in C, F respectively. Then shall the area of the figure bounded by the circumferences of all the semicircles be equal to the area of the circle on CF as diameter.
Proposition 15: Let AB be the diameter of a circle., AC a side of an inscribed regular pentagon, D the middle point of the arc AC. Join CD and produce it to meet BA produced in E; join AC, DB meeting in F, and Draw FM perpendicular to AB. Then
EM = (radius of circle).
Reference
 Great Books of The Western World, v. 11, Encyclopaedia Britannica, 1952
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