This is hard to say. But this may be one. Add the circumcircle to the picture.

For any ΔABC, if

M is the midpoint of the base BC,
L is the foot of the angle bisector at A,
H the foot of the altitude AH

then L is always between M and H. Further, if AL is extended to cross the circumcircle a second time at K, then MK is perpendicular to BC, hence, parallel to AH.

Now, you say that M = L. This implies that MK and AK are one and the same line so that AK is perpendicular to BC, meaning M = L = H.

Having two right angles at M you can now use SAS to claim equality of the two internal triangles.