let $ABCDE$ be a convex pentagon such that

$\angle BAC =\angle CAD =\angle DAE$ and $\angle ABC =\angle ACD =\angle ADE$.

Diagonals $BD$ and $CE$ meet at $P$. Prove that ray $AP$ bisects $CD$.

## ISL 2006 G3

### ISL 2006 G3

The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

- Charles Caleb Colton

### Re: ISL 2006 G3

My solution

The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

- Charles Caleb Colton

- Atonu Roy Chowdhury
**Posts:**63**Joined:**Fri Aug 05, 2016 7:57 pm**Location:**Chittagong, Bangladesh

### Re: ISL 2006 G3

Solution with spiral similarity is easy. Here I'm gonna give a different solu. And I chased some angles and lengths.

This was freedom. Losing all hope was freedom.