The angles at one of the bases of the trapezoid are 86 and 4, and the line segments connecting

The angles at one of the bases of the trapezoid are 86 and 4, and the line segments connecting the midpoints of the opposite sides of the trapezoid are 4 and 1. Find the base of the trapezoid.

Let ABCD be a trapezoid, AD – bases, K and L – midpoints of sides AB and CD, respectively. The sum of the angles at one of the bases is 85 + 5 = 90, so this is the larger AD base.

Extend the sides until they intersect at O.

Angle AOD = 180 – (86 + 4) = 180 – 90 = 90.

N is the middle of AD. Then ОN = AD / 2 is the median of the right-angled triangle AOD.

Since ON bisects any segment with ends on AO and DO of triangle AOD parallel to AD, it intersects BC also in its middle M.

ОМ = BC / 2.

MN = (AD – BC) / 2.

KL = (AD + BC) / 2.

AD = MN + KL.

MN = 4.

KL = 1.

AD = 4 + 1 = 5;

BC = KL – MN.

BC = 4 – 1 = 3.

Solution: 5; 3.



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