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### A cool Geo!

Posted: Wed May 04, 2016 7:54 pm
In $\triangle ABC$, perpendicular bisector of sides $AB$ and $AC$ meet the internal bisector of $\angle BAC$ at $X$ and $Y$, respectively.

Prove that if circle $ACX$ touches $BC$ at $C$ and meets $AB$ again at $Z$ then $BZ=CA$ and circle $ABY$ touches $BC$ at $B$.

### Re: A cool Geo!

Posted: Mon Apr 02, 2018 12:34 am
$$\angle{BAP}=\angle{PAC}=\angle{PCX}$$
Again, $$\angle{PAC}=\angle{YAC}=\angle{YCA}$$
So, $$\angle{ACY}=\angle{PCX}$$
In triangle $ABC$, $CY$ is the isogonal of $CX$
So,
$$\frac{AC^2}{CP^2}=\frac{AY*AX}{YP*XP}$$
Again, $$\frac{AC}{CP}=\frac{AB}{BP}$$
So,
$$\frac{AB^2}{BP^2}=\frac{AY*AX}{YP*XP}$$
It gives us that, $\angle{XBA}=\angle{YBP}$
As, $$\angle{XBA}=\angle{XAB}=\angle{YAB}$$
So, $$\angle{YAB}=\angle{YBP}$$
And we are done.......