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.
∠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
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