MathDB

8.7

Part of 2003 All-Russian Olympiad Regional Round

Problems(1)

BK = 2DC if <ABC =<KAD =<AKD - All-Russian MO 2003 Regional (R4) 8.7

Source:

9/17/2024
In triangle ABCABC, angle CC is a right angle. Found on the side ACAC point DD, and on the segment BDBD, point KK such that ∠ABC=∠KAD=∠AKD\angle ABC = \angle KAD =\angle AKD. Prove that BK=2DCBK = 2DC.
geometryequal segmentsequal angles