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Geometry Difficulty 8.5 Shortlist Prove it Baltic Way

Let $ABCD$ be a trapezoid with $AD \parallel BC$ and $\angle ADC = \angle BAD$ , and let $\ell$ be a line not intersecting the linesegments $AC$ or $BD$ . Assume that $\ell$ intersects the lines $AC, AD, BC, BD$ and $CD$ in the points $P, Q, R, S$ and $T$ respectively. Show that the three circles $\odot(DQT), \odot(BRQ)$ and $\odot(BPS)$ intersect in a common point, where $\odot(XYZ)$ denotes the circumcircle of $XYZ$ .

Solution

Note first that $ABCD$ is concyclic since it is an isosceles trapezoid. We define $X$ to be the intersection of $\odot(DQT)$ and $\odot(ABCD)$ , and it now suffices to prove that $BRQX$ and $BPSX$ are cyclic quadrilaterals. We have that
$\begin{aligned} \angle BXQ &= \angle BXD + \angle DXQ = (\pi - \angle BCD) + \angle DTQ \\ &= \angle RCD + \angle CDR = \angle RCT + \angle CTR = \pi - \angle CRT \end{aligned}$
and hence $BRQX$ is a cyclic quadrilateral. Now observe that
$\begin{aligned} \angle AXQ &= \angle AXD + \angle DXQ = (\pi - \angle ACD) + \angle DTQ \\ &= \angle PCD + \angle CTP = \angle PCT + \angle CTP \\ &= \pi - \angle CPT = \pi - \angle APQ \end{aligned}$
so $APQX$ is concyclic, and it now follows that:
$\angle SBX = \angle DBX = \angle DAX = \angle QAX = \angle QPX = \angle SPX$
Hence, $BPSX$ is a cyclic quadrilateral.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.