## Πέμπτη, 12 Οκτωβρίου 2017

### 2001 JBMO problem 2 (BUL)

proposed by Bulgaria

Let ${ABC }$  be a triangle with ${\angle C = 90^\circ }$  and ${CA \ne CB }$. Let ${CH }$  be an altitude and ${CL }$  be an interior angle bisector. Show that for ${X \ne C }$  on the line ${CL }$, we have ${\angle XAC \ne \angle XBC }$. Also show that for ${Y \ne C }$  on the line ${CH }$  we have ${\angle Y AC \ne \angle YBC }$.

posted in aops here