## ISL 2006 G3

For discussing Olympiad level Geometry Problems
dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### ISL 2006 G3

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

- Charles Caleb Colton

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### Re: ISL 2006 G3

My solution
Let $BD$ and $CD$ meet $AC$ and $AD$ at $X$ and $Y$ resp.

lets define $\angle ACE = \beta$, $\angle DCE = \alpha$, $\angle BDA = \theta$, $\angle CDB = \gamma$

Now, we can easily get that $\bigtriangleup ABD \cong \bigtriangleup ACE$ and $\bigtriangleup BCD \cong \bigtriangleup CDE$

this gives us that $ABCP$ and $APDE$ is concyclic. So, $\angle CBD = \angle CAM = \alpha$ and $\angle DAM = \gamma$. Now.sine law on $\bigtriangleup CDY$ and $\bigtriangleup CYA$ gives us

$\dfrac{DY}{YA} = \dfrac{\sin \alpha \sin (\alpha + \gamma)}{\sin \beta \sin(\gamma + \theta)}$

Again, sine law on $\bigtriangleup CXB$ and $\bigtriangleup ABX$ gives us

$\dfrac{AX}{XC} = \dfrac{\sin \beta \sin(\gamma + \theta)}{\sin \alpha \sin (\alpha + \gamma)}$

Now, let $AP \cap CD = M$. Ceva's theorem on $\bigtriangleup ACD$ gives us $\dfrac{CM}{MD} = 1$, or,
$CM = MD$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

Atonu Roy Chowdhury
Posts: 40
Joined: Fri Aug 05, 2016 7:57 pm
$\triangle AED$ and $\triangle ABC$ are similar and so $AC.AD=AB.AE$ which along with $\angle BAD = \angle CAE$ implies $\triangle ABD$ and $\triangle ACE$ are similar and so $ABCP$,$AEDP$ are cyclic. Again trivial angle chasing yields $\angle CAM = \angle PCM$ where $M$ is the intersection point of $AP$ and $CD$ . So, $\triangle CAM$ and $\triangle PCM$ are similar. That implies $CM^2=AM.PM$. Similarly $DM^2=AM.PM$ .