# The midpoint E of the side AD of a convex quadrangle is equidistant from all its vertices. Find AD if BC = 8

**The midpoint E of the side AD of a convex quadrangle is equidistant from all its vertices. Find AD if BC = 8, and the angles B and C of the quadrangle are 92 ° and 148 ° respectively**

The sum of the angles of any convex n-gon is (n-2) 180, then the sum of the angles of the quadrangle (4-2) 180 = 360.

Those. ∠A + ∠B + ∠C + ∠D = 360

∠A + 92 ° + 148 ° + ∠D = 360 °

∠A + ∠D = 120 °

Triangles AEB, BEC and ECD are isosceles, because side AE = EB = EC = ED.

Hence:

∠A = ∠ABE

∠EBC = ∠ECB

∠ECD = ∠D

I use the sum of the angles of the quadrangle, write:

∠A + ∠ABE + ∠EBC + ∠ECB + ∠ECD + ∠D = 360 °

Using the previously obtained equalities, we write:

∠A + ∠A + 2∠EBC + ∠D + ∠D = 360 °

2∠A + 2∠EBC + 2∠D = 360 °

∠A + ∠EBC + ∠D = 180 °

120 ° + ∠EBC = 180 °

∠EBC = 60 °

Consider the triangle EBC.

By the theorem on the sum of the angles of a triangle, ∠BEC is also equal to 60 °.

Therefore, the triangle EBC is equilateral (by property).

So BC = BE = EC = 8 (by definition) and

8 = BE = EC = AE (by the condition of the problem).

AD = AE + ED = 8 + 8 = 16

Answer: AD = 16