Saudi Booklet

Denote $AH$ as the altitude of triangle $ABC$ then clearly $H \in \Omega_B, \Omega_C$ so $AH$ is the common chord of the two circles. Let $EF$ be the common tangent near $A$ of the two circles with $E \in \Omega_B, F \in \Omega_C$. According to the familiar property, $AH$ passes through the midpoint $K$ of $EF$. Draw $Ax \parallel EF$ then $A(Kx, EF) = -1$.

We redefine $D$ as a point on $BC$ such that $AD \perp EF$. Let $I, J$ be the two incenters of $ABD, ACD$ respectively and let $AI, AJ$ intersect $BC$ at $M, N$ respectively. Let $R, S$ be the centers of the circles $\Omega_B, \Omega_C$ (which are also the midpoints of $AB, AC$) then $RE \parallel AM$ (because they are both perpendicular to $EF$) so
$$\angle ABE = \frac{1}{2}\angle ARE = \frac{1}{2}\angle DAB = \angle MAB,$$
which implies $BE \parallel AM$ or $AE \perp AM$. Similarly, $AF \perp AN$.

Now draw $Ay \parallel MN$ then $AK \perp Ay$. From there, according to the orthogonality property of two harmonic bundles, we immediately have $A(Dy, MN) = -1$. Thus, $DM = DN$. Then, according to the property of the angle bisector,
$$\frac{IM}{IA} = \frac{DM}{DA} = \frac{DN}{DA} = \frac{JN}{JA} \implies IJ \parallel MN.$$
Therefore, the distances from $I, J$ to the line $BC$ are equal implies that the radius of the circle inscribed in triangles $ABD, ACD$ are equal. So in other words, $D$ is the given point in the problem. $\square$