ABC is a given triangle. ABDE and ACFG are the squares drawn outside of the triangle. The points M and N are the midpoints of DG and EF, respectively. Find all the values of the ratio MN : BC.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Sharon Noguchi, SJMN)

Let P, Q, I and J be the midpoints of AC, AB, AD and AF, respectively. We observe the following

QP || BC, QP = BC, IQ || AE, IQ = AE, NJ || AE, NJ = AE, IM || AG, IM = AG, JP || AG, JP = AG.

From there, IQ || NJ and IQ = NJ, IM || PJ and IM = PJ, ∆MIQ = ∆PJN, and we have MQ = NP, and ∠IMQ = ∠JPN (*)

On the other hand, since IM || PJ, ∠IMP = ∠JPM.

Combining with (*), we have ∠QMP = ∠NPM, and with MQ = NP as proved earlier, MNPQ is a parallelogram.

Therefore, MN || QP and MN = QP, or MN : BC = 1 : 2.

For a triangle ABC with obtuse angle BAC, the proof is similar.