## China National Olympiad 2018 P4

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

### China National Olympiad 2018 P4

$ABCD$ is a cyclic quadrilateral whose diagonals intersect at $P$. The circumcircle of $\triangle APD$ meets segment $AB$ at points $A$ and $E$. The circumcircle of $\triangle BPC$ meets segment $AB$ at points $B$ and $F$. Let $I$ and $J$ be the incenters of $\triangle ADE$ and $\triangle BCF$, respectively. Segments $IJ$ and $AC$ meet at $K$. Prove that the points $A,I,K,E$ are cyclic.

This was freedom. Losing all hope was freedom.

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

### Re: China National Olympiad 2018 P4

$Z$ is the intersection point of $XE$ and $YF$.

It is obvious that $X, P, Y$ are collinear.

**Claim 1:**$\triangle AXP$ and $\triangle BYP$ are similar.

Proof: $\angle AXP = \angle ADP = \angle PCB = \angle BYP$

and $\angle APX = \frac {\angle APD}{2} = \frac {\angle BPC}{2} = \angle BPY$

**Claim 2:**$XY||IJ$

Proof: $\frac{XI}{YJ} = \frac{XA}{YB} = \frac{PA}{PB} = \frac{\sin \angle PBA}{\sin \angle PAB} = \frac{\sin \angle PYZ}{\sin \angle PXZ} = \frac{XZ}{YZ}$

Now we get back to our problem.

$\angle IKA = \angle XPA = \frac {\angle DPA}{2} = \frac {\angle DEA}{2} = \angle IEA$

$Q.E.D.$

This was freedom. Losing all hope was freedom.

### Re: China National Olympiad 2018 P4

gettin' headaches seein' this one!