Problem 1783

a) Redefine the point $K$ as the orthocenter of the triangle $BCD$, then $B$ is also the orthocenter of the triangle $KCD$, we will show that $K \in PQ$. Construct the parallelogram $BCDT$.

Since $K$ is the orthocenter of triangle $BCD$, $BC \perp KD$, and $BC \parallel TD$ so $TD \perp KD$. Similarly, $TC \perp KC$ so $KCTD$ is inscribed in a circle of diameter $KT$. Since $CD$ is a common tangent to $(O_1), (O_2)$, $\angle BCD = \angle CAB$, $\angle BDC = \angle DAB$. Therefore
$$180^\circ - \angle CBD = \angle BCD + \angle BDC = \angle CAB + \angle DAB = \angle CAD.$$
Since $BCTD$ is a parallelogram, $\angle CBD = \angle CTD$, hence $180^\circ - \angle CTD = \angle CAD$ entails $ACTD$ internal. Therefore, the points $A, C, T, D, K$ belong to the circle of diameter $KT$. Since $BP, BQ$ are the diameters of $(O_1), (O_2)$, $\angle BAP = \angle BAQ = 90^\circ$, so $A, P, Q$ are collinear. It is left to prove that $KA \perp AB$. From $KACD$ is inscribed, we have $\angle KAC = \angle KDC = 90^\circ - \angle BCD = 90^\circ - \angle BAC$ implies that
$$\angle KAB = \angle KAC + \angle BAC = 90^\circ.$$
Hence $K, P, Q$ are collinear and thus the original problem is solved.

b) Draw the altitude $BH$ of the triangle $BCD$ and $M$ is the midpoint of $CD$. We have
$$\angle KBP = 180^\circ - (\angle CBP + \angle CBH) = 180^\circ - (90^\circ - \angle BPC + 90^\circ - \angle BCH) = 2\angle BCH.$$
Since $BX$ is the bisector of $\angle KBP$, then $\angle KBX = \angle BCM$. We can see that $AKMH$ is cyclic since $\angle KAM = \angle KHM = 90^\circ$ so $\angle BKX = \angle BMC$. This implies that
$$\Delta BKX \sim \Delta CMB \implies \frac{BK}{CM} = \frac{KX}{MB}.$$
Similarly, $\Delta BKY \sim \Delta DMB \implies \frac{BK}{DM} = \frac{KY}{MB}$. And $CM = DM$ so we get
$$\frac{KX}{MB} = \frac{KY}{MB} \implies KX = KY.$$
$\Box$