# The angles at one of the trapezoid bases are 77 ° and 13 °, and the segments connecting the midpoints of the opposite

**The angles at one of the trapezoid bases are 77 ° and 13 °, and the segments connecting the midpoints of the opposite sides of the trapezoid are 11 and 10. Find the trapezoid bases**

Extend sides AB and CD to the intersection at K.

Consider the triangle AKD.

By the theorem on the sum of the angles of a triangle:

∠AKD + ∠KDA + ∠DAK = 180 °

∠AKD + 13 ° + 77 ° = 180 °

∠AKD = 90 °

Therefore, the triangle AKD is rectangular with the hypotenuse AD.

KF is the median (by the condition of the problem).

Mentally describe a circle around this triangle. Since the triangle is rectangular, the center of the circle is located in the middle of the hypotenuse AD (by the theorem on the circumscribed circle).

Therefore, AF = FD = R is the radius of the circle, the median KF is also equal to the radius and, therefore, equal to AD / 2.

Consider the triangle GKH.

For this triangle, KO is the median and is equal to half the hypotenuse GH (as in the previous triangle).

KO = OH = GH / 2

In the triangle BKC – a similar situation: KE = EC = BC / 2

Back to the GKH triangle:

KO = OH = GH / 2 = 11/2 = 5.5

5.5 = OH = KE + EO

KE = EC (we found out earlier)

EO = EF / 2 (since EF is halved by the segment GH by the condition of the problem)

Therefore, we can write:

5.5 = OH = KE + EO = EC + EF / 2

EC = 5.5-EF / 2 = 5.5-10 / 2 = 0.5

BC = 2 * EC = 2 * 0.5 = 1

Consider the trapezoid ABCD.

GH is the midline, therefore GH = (BC + AD) / 2

2GH = BC + AD

AD = 2GH-BC = 2 * 11-1 = 22-1 = 21

Answer: AD = 21, BC = 1