MathDB

P10

Part of 2017 Sharygin Geometry Olympiad

Problems(1)

Circumcenter equidistant from two points

Source: Sharygin Geometry Olympiad Correspondence Round 2017 P-10 (Grade-8-10)

5/1/2017
Points KK and LL on the sides ABAB and BCBC of parallelogram ABCDABCD are such that ∠AKD=∠CLD\angle AKD = \angle CLD. Prove that the circumcenter of triangle BKLBKL is equidistant from AA and CC. Proposed by I.I.Bogdanov
geometrycircumcircle