**ABC is an acute-angled triangle and M is the midpoint of BC. The point P lies in the interior of triangle ABC such that ∠MPB + ∠CAP = 180. N is the midpoint of AP. Then ∠MNP + ∠CPB = 180.**

**Proof**

It is irresistible to let the reflection of P about M be Q. Since M bisects PQ and BC, PBQC is a parallelogram.

Since BP is parallel to QC, ∠PQC = ∠MPB.

So ∠PQC + ∠CAP = 180. Hence PQCA is a cyclic quadrilateral by the converse of ‘opposite angles in a cyclic quad add up to 180’

Since M is the midpoint of PQ and N is the midpoint of PA, triangle PMN is taken to triangle PQA by an enlargement, centre P and scale factor 2; therefore, ∠QAP = ∠MNP.

Also, ∠CPQ = ∠CAQ (same segment).

∠CAQ + ∠QAP = ∠CAP so ∠CPQ + ∠MNP = ∠CAP.

So 180 = ∠MPB + ∠CAP = ∠MPB + ∠CPQ + ∠MNP = ∠CPB + ∠MNP.

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