Instruction

1

The problem of finding an arbitrary angle in

**a trapezoid**requires enough additional data. Consider an example in which two base angle**of a trapezoid**. Let the known angles ∠BAD and ∠CDA, we find the angles ∠ABC and ∠BCD. A-line has a property such that the sum of the angles on each side equal to 180°. Then ∠ABC = 180°-∠BAD and ∠BCD = 180°-∠CDA.

2

Another problem can be the equal sides

**of the trapezoid**and any additional corners. For example, as shown, may be aware that the sides AB, BC and CD are equal, and the diagonal is from lower base angle ∠CAD = α.Consider the triangle**polygon**ABC, it is isosceles as AB = BC. Then ∠BAC = ∠BCA. Let us denote it x for short, and ∠ABC - y. The sum of the angles of any triangle,**angle**a is 180°, it follows that 2x + y = 180°, then y = 180° - 2x. At the same time of the properties**of a trapezoid**: y + x + α = 180° and therefore 180° - 2x + x + α = 180°. Thus, x = α. We found two angles**of a trapezoid**: ∠BAC = 2x = 2α and ∠ABC = y = 180° - 2α.Since AB = CD by hypothesis, then the trapezoid is isosceles or isosceles. So the diagonal are equal and equal angles at the bases. Thus, ∠CDA = 2α and ∠BCD = 180° - 2α.