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Geometry Difficulty 4.6 AIME Find the answer

Let ABCA B C be a triangle, and let D,ED, E, and FF be the midpoints of sides BC,CAB C, C A, and ABA B, respectively. Let the angle bisectors of ∠FDE\angle F D E and ∠FBD\angle F B D meet at PP. Given that ∠BAC=37∘\angle B A C=37^{\circ} and ∠CBA=85∘\angle C B A=85^{\circ}, determine the degree measure of ∠BPD\angle B P D.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Because D,E,FD, E, F are midpoints, we have ABC∼DEFA B C \sim D E F. Furthermore, we know that FD∥ACF D \| A C and DE∥ABD E \| A B, so we have ∠BDF=∠BCA=180−37−85=58∘\angle B D F=\angle B C A=180-37-85=58^{\circ} Also, ∠FDE=∠BAC=37∘\angle F D E=\angle B A C=37^{\circ}. Hence, we have ∠BPD=180∘−∠PBD−∠PDB=180∘−85∘2−(37∘2+58∘)=61∘\angle B P D=180^{\circ}-\angle P B D-\angle P D B=180^{\circ}-\frac{85^{\circ}}{2}-\left(\frac{37^{\circ}}{2}+58^{\circ}\right)=61^{\circ}

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