Secondary Special Camp 2011: Geometry P 1

[Moon]

Problem 1: Let $I$ be the incenter of triangle $ABC$. $O_1$ a circle passing through $B$ and tangent to the line $C I$ at $I$ and $O_2$ a circle passing through $C$ and tangent to the line $BI$ at $I$. Prove that $O_1,O_2$ and the circumcircle of $ABC$ pass through a single point.

[MATHPRITOM]

A VERY EASY PROBELM.

[*Mahi*]

Image:

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[*Mahi*]

$\text {Solution:}$

(Sorry as I have changed the name of a few points.)
We have circle $O_1$ and $O_2$ meeting at $K$.It is therefore enough to prove that point $K$ lies on the circumcircle of $\triangle ABC$
Now,
$ \angle BDC = 180-\angle DBC- \angle DCB$
$=180-\angle \frac{B}{2}-\angle \frac{C}{2}$
$=90+\angle \frac{A}{2}$
Now ,let $MD$ & $OD$ be diameter of circle $O_1$ & $O_2$ respectively.
So,$\angle MDO = 180- \angle BDC $
As $\angle BDO+\angle CDM=180-\angle MDO$
Or,$\angle MDO=90-\angle \frac{A}{2}$
Now,$\angle MDO=180- \angle DMK -\angle DOK$
$=180-\angle DBK -\angle DCK $
$=180-\angle DBC -\angle CBK -\angle DCB -\angle BCK$
So,$180-\angle BDC=\angle BDC -\angle CBK -\angle BCK$
Or,$360-2\angle BDC=180-\angle CBK -\angle BCK$
Or,$2\angle MDO=\angle BKC$
Or,$2(90-\angle \frac{A}{2})=\angle BKC$
Or,$180-\angle A=\angle BKC$
So point $A,B,C.K$ lies on a circle,namely the circumcircle of $\triangle ABC$