It is known that in △ ABC, ∠ BAC = 120 ° with AB and AC as sides, positive △ abd and positive △ ace are made outward respectively, and M is in AD It is known that in △ ABC, ∠ BAC = 120 ° with AB and AC as sides, make positive △ abd and positive △ ace outward respectively, M is the midpoint of AD, n is the midpoint of AE, and P is the midpoint of BC. It is proved that MP = NP

It is known that in △ ABC, ∠ BAC = 120 ° with AB and AC as sides, positive △ abd and positive △ ace are made outward respectively, and M is in AD It is known that in △ ABC, ∠ BAC = 120 ° with AB and AC as sides, make positive △ abd and positive △ ace outward respectively, M is the midpoint of AD, n is the midpoint of AE, and P is the midpoint of BC. It is proved that MP = NP

prove:
Connecting BM and cn
Because △ abd and △ ace are equilateral triangles
Therefore, bad = CAE = 60
Because ∠ BAC = 120
Therefore, ban = cam = 180
So C, a, m are on the same line
Because m is the midpoint of AD
So BM ⊥ ad
So delta BCM is a right triangle
Because P is the midpoint of the hypotenuse BC
So PM is the center line on the hypotenuse
So PM = BC / 2
Similarly, PN = BC / 2
So MP = NP