In the convex quadrilateral ABCD, the angles A and D are equal, and the midpoint

In the convex quadrilateral ABCD, the angles A and D are equal, and the midpoint penpendiculars to the sides AB and CD intersect on the side AD. Prove that AC = BD

Consider a quadrilateral ABCD.

Let’s construct the perpendiculars to the sides AB and CD.

Since they intersect at point O on the AD side, OM and OH are perpendiculars to AB and CD.

By the property of the median perpendicular, we have that:

OA = OB and OC = OD.

Therefore, triangles ABO and CDO are isosceles and angles:

OAB = OBA and OCD = ODC.

By the condition of the problem, we have that OAB = ODC. Hence:

OAB = OBA = OCD = ODC. Hence, AOB = 180 – 2 * OAB = 180 – 2 * ODC = COD.

Then the angle COA = AOB + BOC = COD + BOC = BOD.

By the cosine theorem for the triangle AOC we have:

AC ^ 2 = OA ^ 2 + OC ^ 2 – 2 * OA * OC * cos (COA)

By the cosine theorem for the triangle ABD we have:

BD ^ 2 = OB ^ 2 + OD ^ 2 – 2 * OB * OD * cos (BOD).

It follows from the equations obtained that AC ^ 2 = BD ^ 2 and AC = BD, as required.