In rectangle ABCD, point M-midpoint BC, prove that triangle AMB is isosceles.

If point M is the midpoint of side BC in rectangle ABCD,

then segment BM = segment MC.

The segments BA and CD, being opposite sides of the rectangle, are equal by definition.

Thus, we get two similar triangles ABM and MCD, which are equal to each other:

equality of 2 sides (AB = CD; BM = MC) + 90 ° angle between them.

From the equality of right-angled triangles ABM and MCD follows and the equality of their hypotenuses AM = MD,

hence triangle AMD has two equal sides and is isosceles.