Let $\triangle ABC$ an acute triangle and $O$ the middle of $[BC]$.

Let $\mathcal{C}$ the circle with center in $O$ and radius $OA$.

Let $AB\cap \mathcal{C}=\lbrace D \rbrace$, $AC\cap \mathcal {C}=\lbrace E \rbrace$.

We denote the intersections of the tangents to $\mathcal{C}$ id $D$ an $E$ with $M$.

If $P$ is the middle of $[AM]$ show that $PB=PC$.

First I considered $AO\cap BC=\lbrace Z \rbrace$. After that I draw the parallel to $BC$ trough $Z$ which intersects $AB$ in $X$ and $AC$ in $Y$.

Then $BP$ and $CP$ are middle lines. So it's enough to prove that $MX=MY$.

  • $\begingroup$ I don't think such construction is possible for an acute triangle. Can you provide a picture? $\endgroup$ – Vasya May 22 '18 at 19:25
  • $\begingroup$ Possible duplicate of Show that $OB=OC $ $\endgroup$ – John McClane May 27 '18 at 16:20

Let $F$ be the point diametrically opposite to $A$ on $\mathcal C$. Then $\angle FDB = 90^{\circ}$. Also note that $\angle MDO = 90^{\circ}$. Let $\angle DBF=\alpha$. $BACF$ is a parallelogram because $BO=OC$ and $AO=OF$. Thus, $\angle DAE=\alpha$. Clearly, $\triangle DOM=\triangle EOM$. Thus, $\angle DOM=\frac 1 2 \angle DOE=\angle DAE=\alpha$. We have shown that $\triangle DBF \sim \triangle DOM$. Now consider the rotational homothety with the center in $D$ that sends $M$ to $O$ (its angle is $90^{\circ}$ and its factor is $\cot \alpha$). Clearly, it also sends $F$ to $B$. So it sends the segment $FM$ to $BO$ and $FM \perp BO$. As $OP$ is a midline of $\triangle FAM$, $OP \parallel FM$ and $OP \perp BO.$ Thus, $\triangle BOP=\triangle COP$.


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