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Geometry Difficulty 5.1 AIME, harder Prove it India

Let $ABC$ be an isosceles triangle with $AB = AC$ . Let $D$ be a point on the segment $BC$ such that $BD = 2DC$ . Let $P$ be a point on the segment $AD$ such that $\angle BAC = \angle BPD$ . Prove that $\angle BAC = 2\angle DPC$ .

Solution

Extend $AD$ to $E$ such that $PE = PB$ . Join $EB$ and $EC$ .

$\angle BPE = \angle BAC \text{ and } \frac{PB}{PE} = \frac{AB}{AC} = 1.$

Hence it follows that $CAB$ is similar to $EFB$ . Thus $\angle PEB$ . This shows that $A$ , $B$ , $E$ , $C$ are concyclic. In turn we obtain
$\angle AEC = \angle ABC = \angle ACB = \angle AEB.$

We conclude that $AE$ bisects $\angle BEC$ .

Let $M$ be the mid-point of $BE$ . Join $PM$ and $PC$ . Since $EA$ bisects $\angle BEC$ , we have
$\frac{CE}{EB} = \frac{CD}{DB} = \frac{CD}{2CD} = \frac{1}{2}.$

Thus $CE = EB/2 = EM$ . We also observe that $\angle MEP = \angle CEP$ . This shows that $CEP$ is congruent to $MEP$ . This implies that
$\angle BAC = \angle BPE = 2\angle MPE = 2\angle DPC.$

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