MathDB

1

Part of 2024 India National Olympiad

Problems(1)

Center lies on circumcircle of other

Source: INMO P1

1/21/2024
In triangle ABCABC with CA=CBCA=CB, point EE lies on the circumcircle of ABCABC such that ∠ECB=90∘\angle ECB=90^{\circ}. The line through EE parallel to CBCB intersects CACA in FF and ABAB in GG. Prove that the center of the circumcircle of triangle EGBEGB lies on the circumcircle of triangle ECFECF.
Proposed by Prithwijit De
geometryAngle ChasingINMO