Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Prove it South Africa

Let $ABC$ be an acute-angled triangle with $AB < AC$ , and let points $D$ and $E$ be chosen on the sides $AC$ and $BC$ respectively in such a way that $AD = AE = AB$ . The circumcircle of $ABE$ intersects the line $AC$ at $A$ and $F$ and the line $DE$ at $E$ and $P$ . Prove that $P$ is the circumcentre of $BDF$ .

Solution

Since $AD = AE$ , $AED$ is isosceles, so we also have $\angle ADE = \angle AED$ . Combining this with the fact that $ABPE$ and $ABPF$ are cyclic, we obtain

$\angle ABP = 180^{\circ} - \angle AEP = \angle AED = \angle ADE = \angle ADP$
as well as
$\angle ABP = 180^{\circ} - \angle AFP = \angle DFP,$
so $\angle FDP = \angle DFP$ , which implies that $PD = PF$ . Likewise, $\angle ABE = \angle AEB$ since $AB = AE$ , which means that
$\angle APB = \angle AEB = \angle ABE = \angle APE = \angle APD.$
Now we see that triangles $ABP$ and $ADP$ have the same angles ( $\angle APB = \angle APD$ , $\angle ABP = \angle ADP$ ) and the common side $AP$ , so they are congruent. Thus $PB = PD = PF$ , which means that $P$ is the circumcentre of $BDF$ .

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