PUMaC 2015 · 几何（A 组） · 第 3 题

PUMaC 2015 — Geometry (Division A) — Problem 3

[ 4 ] Cyclic quadrilateral ABCD satisfies ∠ ADC = 2 ⋅ ∠ BAD = 80 and BC = CD . Let the angle bisector of ∠ BCD meet AD at P . What is the measure, in degrees, of ∠ BP D ?

解析

[ 4 ] Cyclic quadrilateral ABCD satisfies ∠ ADC = 2 ⋅ ∠ BAD = 80 and BC = CD . Let the angle bisector of ∠ BCD meet AD at P . What is the measure, in degrees, of ∠ BP D ? Solution: Let S be a point closer to D than C such that ASC is an equilateral triangle and let T be the intersection of the angle bisector of ∠ ACD and the circumcircle ω of ABCD . Note that A is the ◦ 1 1 center of the circumcircle of CDS , so ∠ CSD = ∠ CAD = 10 = ∠ BCD − ∠ BCA = ∠ ACP 2 2 ◦ 1 and ∠ SCD = ∠ SAC − ∠ CSD = 20 = ∠ CAD , so △ AP C ∼ △ CDS . Since AC = CS , 2 △ AP C ≅ △ CDS . ◦ 1 Now, we have ∠ BAD = 40 = ∠ ADC = ∠ T CD since △ CAD is isosceles. Furthermore, since 2 ◦ 1 ABCT is a trapezoid as ∠ ABC = 100 = ∠ BCA + ∠ ACD = ∠ BCT , so AB = CT . Since 2 1 △ AP C ≅ △ CDS , AP = CD . Thus, △ AP B ≅ △ CDT , so ∠ AP B = ∠ CDT = ∠ ACD + 2 ◦ ◦ ∠ ADC = 120 , so ∠ BP D = 60 .