There is incircle $\Gamma$ of triangle $ABC$ tangent to $AB,BC,CA$ respectively at $K,L,M$. Point $D$ is the centre of section $MK$. $|DL|$ is diameter of another circle which intersects with $\Gamma$ at $L,P$ and with $MK$ at $D,R$. Show that line $PR$ divide $AD$ on equal parts.
Actually I have no idea how to prove it, the only thing that that occured to me is to show that AD is a diameter of circle, have anyone idea how to link the line $PR$ with section $AD$ ?
The problem can be re-stated in the following way.
Let $O$ be the circumcenter of the triangle $ABC$, $\Gamma$ the circumcircle. Let $A'$ be the antipode of $A$ in $\Gamma$, $M$ the midpoint of $BC$. The line $A'M$ intersects $\Gamma$ in a point $D$ for which $\widehat{ADM}=\widehat{ADA'}=\frac{\pi}{2}$ clearly holds. Let $E$ be the inverse of $M$ wrt $\Gamma$ and $F$ the midpoint of $ME$. Let $K$ be the projection of $A$ on $BC$.
We want to prove that $D,K,F$ are collinear.
By the Euler's theorem, if we put $O=0$ we have that the orthocenter $H$ of $ABC$ satisfies $H=A+B+C$. Since $A'=-A$, we have that $M=\frac{B+C}{2}$ is the midpoint between $H$ and $A'$. If we call $H'$ the symmetric of $H$ wrt to $BC$, it is well-known that $H'\in\Gamma$. So we have $MH=MH'=MA'$ and $K$ is the midpoint of $HH'$, so, in order to prove our claim, it is sufficient to prove that $D,E,H'$ are collinear. Up to circular inversion (wrt $\Gamma$), this is equivalent to prove that $D,H',M,O$ are concyclic. Finally, this is easy. Since $ADA'$ is a right triangle and $O$ is the midpoint of $AA'$, we have $\widehat{ODA'}=\widehat{OA'M}$. Since $OM$ is perpendicular to $BC$ and $MH'=MA$, $\widehat{MH'O}=\widehat{MA'O}$. $\widehat{MH'O}=\widehat{MDO}$ follows, so $ODH'M$ is a cyclic quadrilateral, QED.
