The midpoint M of the side AD of the convex quadrilateral ABCD is equidistant from all its vertices.

The midpoint M of the side AD of the convex quadrilateral ABCD is equidistant from all its vertices. Find AD if BC = 14 and the angles B and C of the quadrilateral are 110 ° and 100 °, respectively.

By condition, point M is equidistant from all vertices of the quadrangle, and is the midpoint of the segment AD, therefore, a circle of radius AM can be described around the quadrilateral.

Then AM = MD = AC = BM and is the radius of the circle. In a quadrilateral inscribed in a circle, the sum of opposite angles is 180 degrees, then the angle BAM = 180 – BCD = 180 – 100 = 80. Consider a triangle ABM, in which AM = BM, as the radius of a circle, therefore, it is isosceles, then the angle ABM = BAM = 80. Then the angle CBM = ABC – 80 = 110 – 80 = 30.

Let’s draw a perpendicular from point M to the side BC. The BCM triangle is equilateral, therefore point K divides the BC side in half, BK = KС = BC / 2 = 14/2 = 7.

Consider a right-angled triangle BKM, whose leg BK = 7, and the apex angle B = 30. Then the hypotenuse BM = BK / Cos30 = 7 / (√3 / 2) = 14 / √3.

Then side АD = 2 * BК = 28 / √3 = (28 * √3) / 3.

Answer: AD = (28 * √3) / 3.