Suppose ABCD is a trapezium in which AB || CD and AD = BC. Prove that \[\angle A=\angle B\] and \[\angle C=\angle D\].

Hint: Draw a trapezium. Consider 2 triangles in it with the vertex angles as x and y. Prove that x = y. As per the properties of triangles, if 2 angles of 2 triangles are equal then \[{{3}^{rd}}\]angle is also equal.

Complete step-by-step answer:
Consider the trapezium ABCD that is drawn below. It is given that AB || CD.


The sides of trapezium AD are equal to BC, which is marked in the figure.
AD = BC.
We know that trapezium is a quadrilateral whose one pair of opposite sides are parallel to each other. Let us consider that a and b are the length of non-parallel sides and h is the height of the trapezium.
As we are given that AD = BC, thus let’s take side as a.
AD = BC = a.
Let us consider, \[\angle ADC=x\] and \[\angle BCD=y\].
From \[\vartriangle ADM\] (from figure),
\[\sin x=\]opposite side/ hypotenuse\[=\dfrac{AM}{AD}\].
\[\therefore \sin x=\dfrac{h}{a}-(1)\]
Similarly in \[\vartriangle BCN\],
\[\sin y\]= opposite side/ hypotenuse\[=\dfrac{BN}{BC}\].
\[\sin y =\dfrac{h}{a}-(2)\]
From (1) and (2) we can say that,
\[\sin x=\sin y\]
\[\therefore x=y\] i.e. both the angles lie in the \[{{1}^{st}}\]quadrant.
Thus we can say that, \[\angle ADC=\angle BCD\].
 i.e. \[\angle C=\angle D\]
Now, as two angles are the same in two triangles that we have considered, the third angle would also be the same. So we can say that, \[\angle DAB=\angle CBA\].
\[\therefore \angle A=\angle B\].
Hence we proved that \[\angle A=\angle B\] and \[\angle C=\angle D\].

Note: After the construction of trapezium take AM parallel to BN. AM || BN. Similarly, AB || CD, thus AM = BN = h. We have used the property of triangles to solve i.e. we have compared two triangles to prove that the angles subtended are the same.