MathDB
PL^2 = DQ x PQ wanted, DQ = DK, <ADB = 45^o, <KDP = 30^o given

Source: 2009 Balkan Shortlist BMO G1 - easy

April 5, 2020
geometryanglesAngle Chasingangle bisectorequal segments

Problem Statement

In the triangle ABC,∠BACABC, \angle BAC is acute, the angle bisector of ∠BAC\angle BAC meets BCBC at D,KD, K is the foot of the perpendicular from BB to ACAC, and ∠ADB=45o\angle ADB = 45^o. Point PP lies between KK and CC such that ∠KDP=30o\angle KDP = 30^o. Point QQ lies on the ray DPDP such that DQ=DKDQ = DK. The perpendicular at PP to ACAC meets KDKD at LL. Prove that PL2=DQ⋅PQPL^2 = DQ \cdot PQ.
Back to ProblemsView on AoPS