## EGMO 2018 P5

For discussing Olympiad level Geometry Problems
Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm

### EGMO 2018 P5

Let $\tau$ be the circumcircle of triangle $ABC$. A circle $\Omega$ is tangent to the line segment $AB$ and is tangent to $\tau$ at a point lying on the same side of the line $AB$ as $C$. The angle bisector of $\angle{BCA}$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that, $\angle{ABP}=\angle{QBC}$

Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm

### Re: EGMO 2018 P5

Let $N$ be the intersection point of $(ABC)$ and the angle bisector of $\angle{ACB}$ other than $C$
Suppose $\Omega$ touches $\tau$ at $L$ and $AB$ at $M$
Let $M'=NL \cap AB$ and $XY$ be the common tangent of $\tau$ and $\Omega$
Define $K=XY \cap AB$
$KM=KL$
$\measuredangle KLM' = \measuredangle KLN = \measuredangle KLB + \measuredangle BLN = \measuredangle LAB + \measuredangle BAN = \measuredangle LNB + \measuredangle NBA = \measuredangle M'NB + \measuredangle NBM' = \measuredangle NM'B = \measuredangle LM'B = \measuredangle LM'K$
So, we get $KM' = KL$ which means $MM'=0$
By similarity we get $NB^2=NM.NL=NQ.NP$
So, $\triangle NBQ \sim -\triangle NPB$
$\measuredangle NBQ = \measuredangle BPN$
$\measuredangle NBA+ \measuredangle ABQ = \measuredangle BCN + \measuredangle PBC$
$\measuredangle ABQ = \measuredangle PBC$
$\measuredangle ABP = \measuredangle QBC$

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm