Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.3 Shortlist Prove it Turkey

Let $ABCD$ be a cyclic quadrilateral whose sides $BC$ and $AD$ are not parallel. Let $E$ be a point inside the circumcircle of $ABCD$ which is on the opposite side of the line $AB$ with respect to the point $C$ . The lines $DE$ and $AB$ meet at $F$ . Let $G$ be a point inside $ABCD$ and also on the line which is tangent to the circumcircle of triangle $AEF$ at $E$ . If
$\angle GAD = \angle BAF \text{ and } \angle GCB + \angle GBA = \angle EAD + \angle AGD + \angle ABE$
then show that the lines $BC$ , $AD$ and $EG$ are concurrent.

Solution

Let the lines $GE$ and $AB$ meet at a point $M$ . Let $\angle GAD = \angle BAE = a$ , $\angle GAB = b$ . By angle chasing, we have $\angle GED = \angle MEF = \angle MAE = a$ . Therefore, the points $A$ , $E$ , $G$ , $D$ are concyclic. We also get $DGA = \angle DEA = c$ . Let $\angle ABE = d$ . We obtain that $\angle BEG = 180^\circ - 2a - d - c$ . Using the condition given, we get $\angle GCB = 2a + c + d$ . Hence, $\angle BEG + \angle BCG = 180^\circ$ . Hence, we conclude that the points $B$ , $E$ , $G$ , $C$ are concyclic. The line $BC$ is the radical axis of the circles $(ABCD)$ , $(BCGE)$ . The line $EG$ is the radical axis of the circles $(AEGD)$ , $(BCGE)$ . The line $AD$ is the radical axis of the circles $(ABCD)$ and $(AEGD)$ . Therefore, these three lines should be concurrent.

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