Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.9 National Olympiad Prove it Iran

$ABCD$ is a trapezoid with $AB \parallel CD$ . Suppose that the diagonals intersect at $P$ . Let $\omega_1$ be a circle passing through $B$ and tangent to $AC$ at $A$ . Let $\omega_2$ be a circle passing through $C$ and tangent to $BD$ at $D$ . $\omega_3$ is the circumcircle of triangle $BPC$ . Prove that the common chord of circles $\omega_1, \omega_3$ and the common chord of circles $\omega_2, \omega_3$ intersect each other on $AD$ .

Solution

Let $Q$ be the second intersection point of $AD$ with $\omega_1$ , and let $R$ be the second intersection point of $\omega_2$ and $\omega_3$ .

$\left. \begin{array}{l} \widehat{DRB} = 180^\circ - (\widehat{BR} + \widehat{DBR}) \\ \omega_2 : \quad \widehat{BR} = \widehat{RCD} = \theta \\ \omega_3 : \quad \widehat{DBR} = \widehat{PCR} = \beta \end{array} \right\} \implies \widehat{DRB} = 180^\circ - \widehat{ACD}. (\star)$
$AB \parallel CD \implies \left. \begin{array}{l} \widehat{ACD} = \widehat{BAC} = \alpha \\ \widehat{BAC} = \widehat{BQD} = \alpha \end{array} \right\} \underset{\omega_1 :}{\stackrel{\star}{\implies}} \widehat{DRB} = 180^\circ - \widehat{DQB}.$
This implies that the quadrilateral $DQBR$ is cyclic.
So $\widehat{DQR} = \widehat{DBR} = \beta = \widehat{ACR}$ , which means $AQCR$ is also cyclic. Let $CR$ meet $AD$ at $S$ . We have
$\mathcal{P}_{\omega_3}(S) = SR \cdot SC = SA \cdot SQ = \mathcal{P}_{\omega_1}(S).$
(Where $\mathcal{P}_{\lambda}(U)$ is the power of point $U$ with respect to circle $\lambda$ .) That implies $S$ lies on the common chord of $\omega_1, \omega_3$ .
Therefore the common chord of $\omega_1$ and $\omega_3$ , common chord of $\omega_2$ and $\omega_3$ , and the line $AD$ are concurrent at $S$ . ■

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