A cool Geo!
Posted: Wed May 04, 2016 7:54 pm
In $\triangle ABC$, perpendicular bisector of sides $AB$ and $AC$ meet the internal bisector of $\angle BAC $ at $X$ and $Y$, respectively.
Prove that if circle $ACX$ touches $BC$ at $C$ and meets $AB$ again at $Z$ then $BZ=CA$ and circle $ABY$ touches $BC$ at $B$.
Prove that if circle $ACX$ touches $BC$ at $C$ and meets $AB$ again at $Z$ then $BZ=CA$ and circle $ABY$ touches $BC$ at $B$.