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Geometry Difficulty 5.0 AIME Prove it Ukraine

Let MM be the middle point of the side ACAC of triangle ABCABC. Inside △BMC\triangle BMC, there is such point PP that ∠BMP=90∘\angle BMP = 90^\circ, ∠ABC+∠APC=180∘\angle ABC + \angle APC = 180^\circ. Prove that ∠PBM+∠CBM=∠PCA\angle PBM + \angle CBM = \angle PCA.

(Anton Tryhub)

Solution

We construct point B′B' such that ABCB′ABCB' is a parallelogram. Then (Fig. 26)
∠ABC+∠APC=180∘=∠AB′C+∠APC=180∘, \angle ABC + \angle APC = 180^\circ = \angle AB'C + \angle APC = 180^\circ,
hence, quadrilateral APCB′APCB' is inscribed. Obviously, △BPB′\triangle BPB' is isosceles, which yields
∠PCA=∠PB′A=∠PB′M+∠MB′A==∠PBM+∠MBC, Q.E.D. \angle PCA = \angle PB'A = \angle PB'M + \angle MB'A = \\ = \angle PBM + \angle MBC, \text{ Q.E.D.}

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