The median BM is drawn in an isosceles triangle ABC with base AC.

The median BM is drawn in an isosceles triangle ABC with base AC. On the continuation of the median beyond the point M, point D is taken. Prove that the triangles AMD and CMD are equal.

The median BM is drawn in an isosceles triangle ABC with base AC. On the continuation of the median beyond the point M, point D is taken. Prove that the triangles AMD and CMD are equal.

Given: Δ ABC: AB = BC, AC – base, BM – median, D ∈ BM.

Prove: Δ AMD = Δ СMD -?

Proof:

Consider Δ AMD and Δ СMD:

AM = MC (since BM is the median by condition);

MD – common side;

∠ AMD = ∠СMD (since VM is the height).

Therefore, according to the first sign of equality, Δ AMD = Δ CMD. Q.E.D.