# 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 Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.