Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let ABCABC be a triangle, and let DD, EE, and FF be the midpoints of sides BCBC, CACA, and ABAB, respectively. Let the angle bisectors of ∠FDE\angle FDE and ∠FBD\angle FBD meet at PP. Given that ∠BAC=37∘\angle BAC = 37^{\circ} and ∠CBA=85∘\angle CBA = 85^{\circ}, determine the degree measure of ∠BPD\angle BPD.

Solution

Solution:

Answer: 61∘61^{\circ}

Because DD, EE, FF are midpoints, we have ABC∼DEFABC \sim DEF. Furthermore, we know that FD∥ACFD \parallel AC and DE∥ABDE \parallel AB, so we have
∠BDF=∠BCA=180∘−37∘−85∘=58∘ \angle BDF = \angle BCA = 180^{\circ} - 37^{\circ} - 85^{\circ} = 58^{\circ}
Also, ∠FDE=∠BAC=37∘\angle FDE = \angle BAC = 37^{\circ}. Hence, we have
∠BPD=180∘−∠PBD−∠PDB=180∘−85∘2−(37∘2+58∘)=61∘ \angle BPD = 180^{\circ} - \angle PBD - \angle PDB = 180^{\circ} - \frac{85^{\circ}}{2} - \left(\frac{37^{\circ}}{2} + 58^{\circ}\right) = 61^{\circ}

Figure 1

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.