อ้างอิง:
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Problem. Two circles intersect at points $P$ and $Q$. Let $\ell$ be a straight line that intersects the common chord $PQ$. If $\ell$ meets the two circles in the points $A$, $B$, $C$, and $D$ (where $A$, $B$, $C$, and $D$ are arranged in this order on $\ell$), then show that $$\angle APB=\angle CQD\,.$$
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Observe that $\angle APQ$ and $\angle ACQ$ are subtended by the same arc of the circumcircle of the quadrilateral $APCQ$. Consequently, $$\angle ACQ=\angle APQ=\angle APB+\angle BPQ\,.\tag{*}$$
On the other hand, since $\angle ACQ$ is an external angle of the triangle $CDQ$, we get
$$\angle ACQ=\angle CDQ+\angle CQD\,.$$
Since $$\angle CDQ=\angle BDQ=\angle BPQ\,,$$ we conclude that
$$\angle ACQ=\angle BPQ+\angle CQD\,.\tag{#}$$
Hence, from (*) and (#), we get
$$\angle BPQ+\angle CQD=\angle ACQ=\angle APB+\angle BPQ\,.$$
This implies
$$\angle CQD=\angle APB\,.$$