IMO Selection

Solution:

Let $X$ be the tangency point of $CD$ with the first circle and $Y$ the tangency point of $AB$ with the second circle. Further, let $P$ be the intersection of $AC$ with $BD$. As we aim to use Pappus's theorem, we also introduce the points $Q = AX \cap DY$ and $R = BX \cap CY$.

We claim that $\triangle$ AYQ $\sim \triangle$ DXQ $\sim \triangle$ YBR $\sim \triangle$ XCR. Let $\alpha = \angle YAQ$ and $\beta = \angle QYA$. By the tangent chord theorem, $\angle RXC = \alpha$, and as $\angle AXB = 90^\circ$, we have that $\angle DXQ = 90^\circ - \alpha$. Similarly, by the tangent chord theorem, $\angle XCR = \beta$, and as $\angle CYD = 90^\circ$, we have that $\angle QDX = 90^\circ - \beta$. Observe that
$$180^\circ - \alpha - \beta = \angle AQY = \angle XQD = 180^\circ - (90^\circ - \alpha) - (90^\circ - \beta) = \alpha + \beta$$
hence $\beta = 90^\circ - \alpha$. The claim follows immediately.

It now follows from the similarities that $Q$ and $R$ are on the power line of the two circles, as $QA \cdot QX = QY \cdot QD$ and $RY \cdot RC = RB \cdot RX$. By Pappus's theorem, $Q$, $P$, and $R$ are collinear, so $P$ also lies on the power line. We are now done, as the line through the intersection points of the circles is always their power line.