Japan competition problems, 2015

Let $\alpha = \frac{1}{2} \angle CAB$, $\beta = \frac{1}{2} \angle ABC$, $\gamma = \frac{1}{2} \angle BCA$. Since $P$ lies on the arc $AC$ opposite to $B$, we have $\angle APB = \angle ACB = 2\gamma$. We will show that $\angle APE = \alpha + \gamma$ holds, separating cases depending on the positions of the points involved.

* When $BA < BC$ is satisfied:
Since we have
$$\angle BPE = 180^\circ - \angle BIE = 180^\circ - \angle AIB - \angle AIE = 180^\circ - (90^\circ + \gamma) - (\beta + \gamma) = \alpha - \gamma,$$
we have $\angle APE = \angle APB + \angle BPE = 2\gamma + (\alpha - \gamma) = \alpha + \gamma$.

* When $BA > BC$ is satisfied:
Since we have
$$\angle BPE = 180^\circ - \angle BIE = 180^\circ - \angle CIB - \angle CIE = 180^\circ - (90^\circ + \alpha) - (\beta + \alpha) = \gamma - \alpha,$$
we have $\angle APE = \angle APB - \angle BPE = 2\gamma - (\gamma - \alpha) = \alpha + \gamma$.

Thus, in both cases we have $\angle APE = \alpha + \gamma$. From
$$\angle APC = 180^\circ - \angle ABC = 2\alpha + 2\gamma = 2\angle APE,$$
the half-line $PE$ bisects the angle $\angle APC$. Consequently, if we let $M$ be the mid-point of the arc $AC$ (containing the point $B$) of the circle $\Gamma$, then three points $P$, $E$, $M$ lie on the same straight line. Similarly, if we let $N$ be the mid-point of the arc $AB$ (containing the point $C$) of $\Gamma$, then three points $Q$, $D$, $N$ lie on the same straight line.

Let $\Omega$ be the in-circle of the triangle $ABC$, $\ell_1$ be the line tangent to $\Gamma$ at $M$, and $\ell_2$ be the line tangent to $\Gamma$ at $N$. Then, $\ell_1$ is parallel to line $AC$ and $\ell_2$ is parallel to line $AB$. Therefore, if we let $X$ be the point of intersection of $\ell_1$ and $\ell_2$, then the bisector of $\angle BAC$ and the bisector of $\angle MXN$ are parallel. From $AD = AE$ it follows that line $DE$ is perpendicular to the bisector of $\angle BAC$, and

from $XM = XN$ follows that line $MN$ is perpendicular to the bisector of $\angle MXN$. Consequently, lines $DE$ and $MN$ are parallel. Let $O$ be the point of intersection of lines $PE$ and $QD$, then $OE:OD:OM:ON$ holds. By the theorem on the power of a point with respect to a circle, we also have $OP \cdot OM = OQ \cdot ON$, we obtain $OP \cdot OE = OQ \cdot OD$. Points $O$, $E$, $P$ lie on a straight line in this order, and points $O$, $D$, $Q$ lie on a straight line in this order. Hence by the converse part of the theorem on a power of a point with respect to a circle, we conclude that four points $D$, $E$, $P$, $Q$ are con-cyclic.