Problem 1978

Solution 1. Since $ABCD$ has a circumscribed circle, we have $\angle DAQ = \angle DCB$. Also, since the line $AC$ is tangent to the circle $AQD$, we get $\angle CBD = \angle CAD = \angle AQD$. Therefore triangles $ADQ$ and $CDB$ are similar (AA).
Let $R$ be the midpoint of segment $CD$. Using the similarity of triangles above, $N$ and $R$ are corresponding points. Hence we have
$\angle QNA = \angle BRC$.

Let point $K$ be the second intersection of line $CD$ with circle $ABR$ (if $CD$ meets circle $ABR$ at two points, take $K$ to be the other intersection point different from $R$; if $CD$ is tangent to circle $ABR$, take $K = R$). Then we have $\angle BAK = \angle BRC = \angle QNA$; this equality shows that line $AK$ is tangent to circle $ANC$. The same method can be used to prove that line $BK$ is tangent to circle $BMP$. This completes the proof. □

Solution 2. Below we give a proof that does not require the condition that $ABCD$ has a circumscribed circle. As in Solution 1, let points $M$, $N$ lie on lines $BC$, $AD$ respectively, satisfying $BM : MC = DN : NA$.

Let line $AB$ meet $CD$ at point $T$ (if the two lines are parallel, take $T$ to be the point at infinity in that direction). Let line $CD$ meet the tangent to circle $ANQ$ at $A$ at point $K_1$, and let line $CD$ meet the tangent to circle $BMP$ at $B$ at point $K_2$.

Let the points at infinity in the directions of $AD$ and $BC$ be points $I$, $J$ respectively. Note that since $\angle K_1AD = \angle AQN$, $\angle CAD = \angle AQD$, and $\angle IAT = \angle IQT$, the pencils of lines ($AD$, $AC$, $AT$, $AK_1$) and ($QA$, $QD$, $QI$, $QN$) are similar (perspective in angle). Hence we have the following cross ratio:
$$(D, C; T, K_1) = (AD, AC; AT, AK_1) = (QA, QD; QI, QN) = (A, D; I, N) = \frac{DN}{NA}.$$
Similarly we obtain
$$(D, C; T, K_2) = (BD, BC; BT, BK_2) = (PC, PB; PJ, PM) = (C, B; J, M) = \frac{BM}{MC}.$$
Since we already have $BM:MC = DN:NA$ from the start, we get $(D,C;T,K_1) = (D,C;T,K_2)$, that is, $K_1 = K_2$. $\square$