Solution #15

First note that triangles BCM and BPM are both right, because CM is chosen perpendicular to BK. Because BK is the angle bisector of angle B, BCM and BPM are similar. Because they share side BM, they are congruent.

Thus M is the midpoint of CP, and BC = MP = 120.

Similarly, triangles AQN and ACN are congruent. Thus N is the midpoint of CQ, and AQ = AC = 117.

Because M and N are respectively the midpoints of PC and QC, by considering triangle PCQ we see that MN is half of PQ.

AQ + BP = AB + PQ, so PQ = AQ + BP - AB = 120 + 117 - 125 = 112. Thus MN = 56.