Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Argentina

Let $P$ be a point in the exterior of a circumference $\Gamma$ , and let $PA$ be one of the tangents from $P$ to $\Gamma$ . The line $l$ passes through $P$ and intersects $\Gamma$ in $B$ and $C$ , with $B$ between $P$ and $C$ . Let $D$ be the point symmetric to $B$ with respect to $P$ . Let $\omega_1$ and $\omega_2$ be the circumferences circumscribed to the triangles $DAC$ and $PAB$ respectively; $\omega_1$ and $\omega_2$ intersect in $E \neq A$ . The line $EB$ intersects $\omega_1$ in another point $F$ . Prove that $CF = AB$ .

Solution

Note that $E\hat{B}P = E\hat{A}P = E\hat{A}D = E\hat{F}D$ ; then, $l \parallel DF$ . As a consequence, if $K$ is the second intersection point of the line $BC$ with $\omega_1$ , it follows that $CFDK$ is an isosceles trapezoid and so, $CF = KD$ .

On the other hand, since $K\hat{D}A = K\hat{C}A = B\hat{C}A = B\hat{A}D$ , we have that $KD \parallel BA$ , and taking into account that $PA = PD$ , it follows that $KDBA$ is a parallelogram. Then, $KD = AB$ .

Therefore, $CF = KD = AB$ , as we wanted to prove.

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