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Plane Geometry - Bruce E. Shapiro

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SECTION 51. THE POINCARE DISK MODEL 301that d(A, B) = 0. Then(AP )(BQ)∣ln (BP )(AQ) ∣ = 0(BP )(AQ)=⇒(AP )(BQ) = 1=⇒ (BP )(AQ) = (AP )(BQ)=⇒ BPBQ = APAQSuppose A ≠ B. Let C be the intersection of AQ and BP . If they do notintersect, interchange the labels on A and B so that they do intersect.By the vertical angles theorem ∠ACP ∼ = ∠BCQ.The power of point C is independent of the line chosen, henceso thatPower(C) = (CA)(CQ) = (CB)(CP )CACP = CBCQBy the SAS criterion for triangle similarity,△ACP ∼ △BCQHenceTherefore by SAS again,∠P AC ∼ = ∠QBC△P AQ ∼ △P BQBy the converse to the similar triangles theorem,BQAQ = BPAP = P QQP = 1Hence BQ = AQ and BP = AP . Thus A = B.If the points fall on a diameter, then suppose that P ∗ A ∗ B ∗ Q. Then(BP )(AQ) = (AP )(BQ)(P A + AB)(AB + BQ) = (AP )(BQ)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

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