Let $ ABC $ an acute triangle and $O$ is the middle of $[BC] $ .
Let $\mathcal {C} $ the circle with center in $O $ and radius $OA $.
Let $AB\cap \mathcal{C}=\lbrace D \rbrace$ and $AC\cap \mathcal {C} =\lbrace E\rbrace $.
If $ M $ is the intersections of the tangents to the circle $\mathcal {C} $ in $D $ respectively $E $, and $P $ is the middle of $AM $ show that $PB=PC $.
My ideas: First time I tried to make an inversion with respect to the circle $\mathcal {C} $.
The second time I considered $AO\cap \mathcal {C}=\lbrace Z\rbrace $ and I draw $XY $ trough $Z $ s.t. $XY||BC $ and $X\in AB $, $Y\in AC $.
$B $ is the middle of $AX $, $C $ is the middle of $AY $.
Now I have to show that $MX=MY $. I am stuck.
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