, 2019

Let $N$ and $M$ be the midpoints of the $\text{arcs B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\angle F A E = \angle B A C = \angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\varphi$ (dilation in case of $A B = A C$) that moves $A F I E$ to $N B M C$. The directed angle in which $\varphi$ changes directions is $\angle(A F, N B)$, same as $\angle(A P, N P)$ and $\angle(A Q, N Q)$; so lines $A P$ and $A Q$ are mapped to lines $N P$ and $N Q$, respectively. Line $E F$ is mapped to $B C$; we can see that the intersection points $P = E F \cap A P$ and $Q = E F \cap A Q$ are mapped to points $B C \cap N P$ and $B C \cap N Q$, respectively. Denote these points by $P'$ and $Q'$, respectively.

Let $L$ be the midpoint of $B C$. We claim that points $P, Q, D, L$ are concyclic (if $D = L$ then line $B C$ is tangent to circle $P Q D$). Let $P Q$ and $B C$ meet at $Z$. By applying Menelaus' theorem to triangle $A B C$ and line $E F Z$, we have
$$\frac{B D}{D C} = \frac{B F}{F A} \cdot \frac{A E}{E C} = -\frac{B Z}{Z C},$$
so the pairs $B, C$ and $D, Z$ are harmonic. It is well-known that this implies $Z B \cdot Z C = Z D \cdot Z L$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $B C$, because the cross-ratio is preserved.) Hence, $Z D \cdot Z L = Z B \cdot Z C = Z P \cdot Z Q$ by the power of $Z$ with respect to the circumcircle; this proves our claim.

By $\angle M P P' = \angle M Q Q' = \angle M L P' = \angle M L Q' = 90^{\circ}$, the quadrilaterals $M L P P'$ and $M L Q Q'$ are cyclic. Then the problem statement follows by
$$\begin{aligned} \angle D P A + \angle A Q D & = 360^{\circ} - \angle P A Q - \angle Q D P = 360^{\circ} - \angle P N Q - \angle Q L P \\ & = \angle L P N + \angle N Q L = \angle P' M L + \angle L M Q' = \angle P' M Q' = \angle P I Q \end{aligned} .$$

Define the point $M$ and the same spiral similarity $\varphi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F, E$ to $B, C$ is the Miquel point of the lines $F E, B C, B F$ and $C E$; that is, the second intersection of circles $A B C$ and $A E F$. Denote that point by $S$.

By $\varphi(F) = B$ and $\varphi(E) = C$ the triangles $S B F$ and $S C E$ are similar, so we have
$$\frac{S B}{S C} = \frac{B F}{C E} = \frac{B D}{C D} .$$
By the converse of the angle bisector theorem, that indicates that line $S D$ bisects $\angle B S C$ and hence passes through $M$.

Let $K$ be the intersection point of lines $E F$ and $S I$. Notice that $\varphi$ sends points $S, F, E, I$ to $S, B, C, M$, so $\varphi(K) = \varphi(F E \cap S I) = B C \cap S M = D$. By $\varphi(I) = M$, we have $K D \parallel I M$.

We claim that triangles $S P I$ and $S D Q$ are similar, and so are triangles $S P D$ and $S I Q$. Let ray $S I$ meet the circumcircle again at $L$. Note that the segment $E F$ is perpendicular to the angle bisector $A M$. Then by $\angle A M L = \angle A S L = \angle A S I = 90^{\circ}$, we have $M L \parallel P Q$. Hence, $\widehat{P L} = \widehat{M Q}$ and therefore $\angle P S L = \angle M S Q = \angle D S Q$. By $\angle Q P S = \angle Q M S$, the triangles $S P K$ and $S M Q$ are similar. Finally,
$$\frac{S P}{S I} = \frac{S P}{S K} \cdot \frac{S K}{S I} = \frac{S M}{S Q} \cdot \frac{S D}{S M} = \frac{S D}{S Q}$$
shows that triangles $S P I$ and $S D Q$ are similar. The second part of the claim can be proved analogously.

Now the problem statement can be proved by
$$\angle D P A + \angle A Q D = \angle D P S + \angle S Q D = \angle Q I S + \angle S I P = \angle Q I P .$$

Denote the circumcircle of triangle $A B C$ by $\Gamma$, and let rays $P D$ and $Q D$ meet $\Gamma$ again at $V$ and $U$, respectively. We will show that $A U \perp I P$ and $A V \perp I Q$. Then the problem statement will follow as
$$\angle D P A + \angle A Q D = \angle V U A + \angle A V U = 180^{\circ} - \angle U A V = \angle Q I P .$$

Let $M$ be the midpoint of $\operatorname{arc}\ \widehat{B U V C}$ and let $N$ be the midpoint of $\operatorname{arc}\ \widehat{C A B}$; the lines $A I M$ and $A N$ being the internal and external bisectors of angle $B A C$, respectively, are perpendicular. Let the tangents drawn to $\Gamma$ at $B$ and $C$ meet at $R$; let line $P Q$ meet $A U, A I, A V$ and $B C$ at $X, T, Y$ and $Z$, respectively.

As in Solution 1, we observe that the pairs $B, C$ and $D, Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B, C$ and $U, P$ are also harmonic. Analogously, the pair $V, Q$ is harmonic with $B, C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\Gamma$ by its harmonic pair with respect to $B, C$. In particular, the inversion maps points $B, C, N, U, V$ to points $B, C, M, P, Q$, respectively.

Combine the inversion with projecting $\Gamma$ from $A$ to line $P Q$; the points $B, C, M, P, Q$ are projected to $F, E, T, P, Q$, respectively.

The combination of these two transformations is projective map from the lines $A B, A C$, $A N, A U, A V$ to $I F, I E, I T, I P, I Q$, respectively. On the other hand, we have $A B \perp I F$, $A C \perp I E$ and $A N \perp A T$, so the corresponding lines in these two pencils are perpendicular. This proves $A U \perp I P$ and $A V \perp I Q$, and hence completes the solution.