Let $N$ and $Y$ be the midpoint of the arc $\text{BAC}$ and the intersection of the lines $ID, T_bN$, respectively. Now, since $\text{YT bA = AN 2 = B-C 2 = AIY}$, therefore $AYT_bI$ is cyclic, so $\angle AYD = \angle AT_bI = \angle C$. Thus, $AYCD$ is cyclic and $AYX_bC$ is a rectangle, so $AYT_b \cong CX_bT_b$ and $\angle CT_bX_b = \angle AT_bY = \frac{B-C}{2}$. Similarly $\angle BT_cX_c = \frac{B-C}{2}$. By easy angle chasing, we get $BT_cX_c \sim CT_bX_b$ which completes the proof.

Suppose the feet of the perpendiculars from the vertex $A$ to the lines $BC, CD$, and $DA$ are respectively $X, Y, Z$.
The quadrilaterals $ABXY$ and $ADZC$ are cyclic, and $AC$ is a diagonal of both. Thus we have:
$$\angle ABX = \angle ACX, \quad \angle ADZ = \angle ACZ$$
Now, since the quadrilaterals are cyclic (respectively with circles of diameter $AC$), we also have:
$$\angle ABX + \angle ADZ = 90^\circ$$
Moreover, since $X$ and $Y$ lie on the angle bisector of $\angle ABC$, and $Z$ and $W$ lie on the angle bisector of $\angle ADC$, therefore:
$$\angle XAY = \angle ZAW$$
So we conclude:
$$\angle XAY = \angle ZAW = 45^\circ$$
It suffices to prove the following; If $D, E, F$ are the points where the incircle touches the sides $BC, CA, AB$ respectively, then by Iran's lemma, we know that $D, E, F$ are collinear. Also, we know that $X$ and $Y$ both lie on the angle bisector of $\angle ABC$, and $Z$ and $W$ on that of $\angle ADC$. Therefore:
$$XY \parallel ZW$$
it can now be concluded that the quadrilateral is an isosceles trapezoid. ■