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Slovenia — National Math Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Slovenia

Let $P$ be a point inside the triangle $ABC$ , such that the angles $\angle CBP$ and $\angle PAC$ are equal. Denote the intersection of the line $AP$ and the segment $BC$ by $D$ , and the intersection of the line $BP$ with the segment $AC$ by $E$ . The circumcircles of the triangles $ADC$ and $BEC$ meet at $C$ and $F$ . Show that the line $CP$ bisects the angle $DFE$ .

Solution

Using the equalities between the inscribed angles in the quadrilaterals $AFDC$ and $BCEF$ as well as the given equality $\angle CBP = \angle PAC$ we find that
$\begin{aligned} \angle CFE &= \angle CBE = \angle CBP = \angle PAC \\ &= \angle DAC = \angle DFC. \end{aligned}$

In the cyclic quadrilateral $BCEF$ we have $\angle EBF = \angle ECF$ and in the cyclic quadrilateral $AFDC$ we have $\angle ACF = \angle ADF$ . So,
$\angle PBF = \angle EBF = \angle ECF = \angle ACF = \angle ADF = \angle PDF.$

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