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Turkey — Team Selection Test

Geometry Difficulty 7.5 National Olympiad, round 2 Prove it Turkey

$ABCD$ is a convex quadrilateral with $\angle ABC = \angle BCD$ . The lines $AD$ and $BC$ intersect at $P$ , and the line passing through $P$ and parallel to $AB$ intersects the line $BD$ at $T$ . Show that $\angle ACB = \angle PCT$ .

Solution

Let the line passing through $A$ and parallel to $BC$ intersect the line $CD$ at $E$ . Since
$\frac{EA}{CP} = \frac{AD}{PD} = \frac{AB}{PT} \quad \text{and} \quad \angle EAB = \angle CPT,$
one has $EAB \sim CPT$ , hence $\angle PCT = \angle AEB$ . On the other hand, one has $\angle ABC = \angle BCD$ , thus $\angle AEB = \angle ACB$ .

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