Geometry problem - prove that incentre of two triangles coincide

Question

$AB$ and $AC$ are tangent to the circle at $B$ and $C$ respectively. Let $OA$ and $BC$ intersect at M. Let $E$ be an arbitrary point on the circle. Extend $EM$ to meet the circle at F.
(1) Prove that $\Delta AEF$ and $\Delta ABC$ have a common incentre.
(2) Prove that $\Delta AEF$ and $\Delta ABC$ have a common excentre.

I verified (1) on Geogebra and one of the interesting things I found was that $\Delta AEM \sim \Delta AOF$, but I have no idea how to prove it (or will it even be useful).

My Question

I want to know how to solve (1). To do it I tried to show that $OA$ is the angle bisector of $\angle MAF$ first. However, although I was quite sure that it is true, I couldn't come up with a proof of it. Anyone can give me a hint?

Answer 1

Draw $AO$, $OE$, $OF$, $OB$ & $OC$. $OB\perp AB$, $OC\perp AC$ & $AO$ will pass through $M$.

Notice that, $EM\cdot MF=BM\cdot MC$.

Since, $BACO$ is a cyclic quadrilateral, $BM\cdot MC=AM\cdot MO$.

Thus, $EM\cdot MF=AM\cdot MO$. $\Rightarrow$ $AEOF$ is a cyclic quadrilateral.

$\angle OAE=\angle EFO=\angle FEO$($\triangle OEF$ is isosceles)$=\angle FAO$

$\Rightarrow$ $AO$ bisects $\angle EAF$.

But, $AO$ bisects $\angle BAC$ as well; Which indicates that the incentres of both $\triangle ABC$ & $\triangle FAE$ lie on $AO$.

Assume that, point $I$ is the incentre of $\triangle ABC$ [ We don't necessarily know whether $I$ is the point of intersection of $AO$ with the circle or not ]. Draw $CI$ & $EI$.

$\Rightarrow \frac{AI}{IM}=\frac{AC}{CM}$[By the angle bisector theorem]$=\frac{CO}{OM}$[$\triangle ACM\sim\triangle COM$]$=\frac{OF}{OM}=\frac{AE}{EM}$[ Since $AEOF$ is a cyclic quadrilateral ]

$\Rightarrow \frac{AI}{IM}=\frac{AE}{EM}$

Hence, $IE$ is the angle bisector of $\angle AEF$ as well; Which indicates that $\triangle ABC$ and $\triangle AEF$ have the same incentre.

Answer 2

First of all, note that $I_{ABC}\equiv$ Midpoint of the arc BC. That's mainly because this incenter shall be on $AO$ and on the line that connects $B$ to the midpoint of the arc $BC$ (because it is the bisector of the angle $ABC$) So we just have to prove that the incenter of $AEF$ lies on the midpoint of arc $BC$. For proving this note that $AE.AF=AC^2$ and thus if we name the intersection of $AF$ and the circle as $T$, $AT.AF=AC^2$ and so $AT=AE$. This shows (in the same time) that the midpoint of the arc $BC$ lies on the bisector of the angles $AFE$ and $FAE$ and hence we are done.