Point M is marked inside the triangle ABC, and on the side of the BC-points D and E so that DM is parallel to AB

Point M is marked inside the triangle ABC, and on the side of the BC-points D and E so that DM is parallel to AB, EM is parallel to AC, BD = DM and CE = EM. prove that the point M is equidistant from the sides of the triangle ABC

Since, according to the condition, BD = DM, then the triangle BDM is isosceles, therefore, the angle DBM = DMB.

The segment AB is parallel to DM, then the angle ABM = BMD as criss-crossing angles at the intersection of parallel AB and DM of the secant BM. Then the angle ABM = DBM, which means BM is the bisector of the angle ABC.

Similarly, triangle MEC is isosceles, since ME = CE, angle EMC = ECM. ME || AC, then the angle ACM = EMC = ECM, and CM is the bisector of the angle ACB.

Then the point M is the point of intersection of the bisectors and is the center of the inscribed circle, which means that the distance from point M to the sides of the triangle is equal to the radius of this circle, which was required to prove.