Denote the common tangent points $T_2$ and $T_1$ to circles $\Gamma$ and $\Omega$ respectively on the side opposite to $B$, and similarly for $T_3$ and $T_4$ Obviously, $IT_2PT_4$ is a cyclic quadrilateral due to $\angle IT_2 = \angle IT_4 = \pi/2$, and $T_2CBT_4$ is a cyclic quadrilateral due to symmetry. We will now prove $IT_2CT_4$ is cyclic, which would imply $IT_2CBT_4$ cyclic and then $ICPB$ cyclic, which would mean $\angle IBP = \pi - \angle ICP$, but due to symmetry these angles are equal and therefor $\pi/2$ Consider the line through $I$ parallel to $BC$ and call this line $l$. Due to tangents to $\Gamma$ at $T_2$ and $F$ being parallel, the points $F$ and $T_2$ are diametrically opposite, and by symmetry through $AI$ we get that $T_2$ reflects to $E$ through $l$. Similarly we get that $T_4$ reflects to $F$ over $l$. We now aim to prove $C$ reflects to $T_1$ through $l$ and same for the other side. First we prove $T_1$, $I$ and $B$ are collinear. Taking a few simple homotheties and reflections we get $\angle T_1IA = \angle T_3DX = \angle FXD = \frac{1}{2} \angle FID = \angle BID = \pi - \angle AIB$, implying $T_1$, $I$ and $B$ are collinear. We use this to finish out reflection claim. It is easy to see that the points $A$, $P$ and intersection points of the sides and tangents form a parallelogram, and $I$ is the center of inscribed circle to this parallelogram so it is the intersection of the diagonals, and therefor the intersection points mentioned earlier lie on $l$. We see that $T_1$ is the intersection point of $BI$ and $T_1T_2$, but the reflection over $l$ of these lines are $IC$ and $AC$ and therefor $T_1$ reflects to $C$, which is what we wanted to prove Now we prove the main claim. By symmetries we have just proved, we have $\angle T_4BT_3 = \angle FT_3E = \pi - \angle EIF = \pi - \angle T_4IT_2$, which means $BT_4IT_2$ is cyclic and by the second paragraph implies the main claim $\blacksquare$

This is nice problem. I see an extension as follows Let $ABC$ be a triangle with $AB=AC$. $(I)$ is incircle of $ABC$. $J$ is exsimilicenter of $(I)$ and circle diameter $AI$. $L$ is isogonal conjugate of $J$ with respect to $ABC$. Prove that tangents from $L$ to $(I)$ are parallel to sides $AB$, $AC$.

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Blastzit wrote: Let $\Delta ABC$ be an isosceles triangle with $AB=AC$. The incircle $\Gamma$ of $\Delta ABC$ has centre $I$, and it is tangent to the sides $AB$ and $AC$ at $F$ and $E$ respectively. Let $\Omega$ be the circumcircle of $\Delta AFE$. The two external common tangents of $\Gamma$ and $\Omega$ intersect at a point $P$. If one of these external common tangents is parallel to $AC$, prove that $\angle PBI=90^{\circ}$. $\textbf{LEMMA:-}$ Let $\omega$ and $\omega_1$ be two circles such that $\omega_1$ passes through the center of $\omega$. Let their common tangents meet at $A$ and be tangent to $\omega_1$ at points $\{B,C\}$. Then $BC$ is tangent to $\omega$. Let $\{O,O_1\}$ be the circumcenters of $\{\omega,\omega_1\}$ respectively. Let $AO_1\cap BC=T$. and let the commom tangents be tangent to $\omega$ at $M,N$ such that $\overline{A-M-B}$ and $\overline{A-N-C}$. So now notice that $\angle AO_1B=MBT$, $\angle BOO_1=\angle BMT\implies \Delta MBT\stackrel{+}\sim\Delta BO_1O\implies BM=MT\implies OM=OT\implies \omega$ is tangent to $BC$. ____________________________________________________________________________________________ From $\textbf{LEMMA}$ we get that if $\overline{AI}\cap\odot(I)=X$ and $\{U,V\}$ are the common Tangency Points of $\odot(AI)$ with the External Common tangents then $\overline{U-V-X}$ Moreover it is tangent to $\odot(I)$. So clearly $I$ is the Incenter of $\Delta UVP$ also $\angle IVA=\frac{\pi}{2}$. So, $(A,I;X,P)=-1$. Now if $\ell_1,\ell_2$ are the common tangents of $\odot(I)\sim\odot(AI)$ then the Quadrilateral formed by the lines $\{\overline{AB},\overline{AC},\ell_1,\ell_2\}$ is a Rhombus as the Inscribed Conic is a circle with $I$ as the center. Consider the Involution $\Psi:K\in\overline{AI}\leftrightarrow K^*\in\overline{AI}$ such that $IK=IK^*$. So, $-1=(A,I;X,P)\overset{\Psi}{=}(P,I;Y,A)\implies\overline{BI}\perp\overline{PB}$. $\blacksquare$