Question

ABCD is a quadrilateral in which \[AB\,||\,CD\]and $AD = BC$, prove that $\angle A = \angle B$.

Answer 1

Hint: In this question, first we will draw a perpendicular line from one parallel side to the other. Then we will prove the congruence of two triangles formed so that we can prove the equality of angles formed by the perpendicular line and the non-parallel line and then we will add the angle of ninety degrees in both the angles to conclude the result.

Complete step by step answer:
In the above question, it is given that \[AB\,||\,CD\] and $AD = BC$and we have to prove that $\angle A = \angle B$.

We know that the sum angle A and angle D is \[{180^ \circ }\].Therefore,
$\angle A + \angle D = {180^ \circ }$.........(angles on the same side of traversal)
Similarly.
$\angle B + \angle C = {180^ \circ }$........(angles on the same side of traversal)
Now we will prove that $\vartriangle ADE \cong \,\vartriangle BCF$,
So, In $\vartriangle ADE\,\,and\,\,\vartriangle BCF$
$AE\, = \,BF$(perpendicular to the same line)
$AD\, = \,BC$(given)
$\angle AED\, = \,\angle BFC\, = \,{90^ \circ }$
Therefore, $\vartriangle ADE \cong \,\vartriangle BCF$.
Hence, $\angle DAE = \angle FBC$ by corresponding parts of the congruent triangle.

We know that $\angle EAB = \angle ABF = {90^ \circ }$
So, from above equation, we know that
$\angle DAE = \angle FBC$
Now adding both sides ${90^ \circ }$.
$ \Rightarrow \angle DAE + {90^ \circ } = \angle FBC + {90^ \circ }$
Also, we have $\angle EAB = \angle ABF = {90^ \circ }$.
Therefore, on substitution we get
$ \Rightarrow \angle DAE + \angle EAB = \angle FBC + \angle ABF$
$ \Rightarrow \angle DAB = \angle ABC$
Therefore, $\angle A = \angle B$
Hence proved.

Note:In the above question, we can also prove that $\angle D = \angle C$ and $DE = FC$. Here we also know that $AB = EF$and ABEF is a square. With the help of conditions given in this question, we can say that the above figure is isosceles trapezium in which one pair of opposite sides are equal and another pair of parallel sides are unequal.