Question

In a quadrilateral ABCD, AB=CD and AD=BC. Prove that ABCD is a parallelogram.

Answer 1

Hint: Start by drawing the diagram, followed by proving that the opposite angles of the quadrilateral ABCD is equal using the properties of congruent triangles. If opposite angles of a quadrilateral are equal, then the quadrilateral is a parallelogram.

Complete step-by-step answer:
Let us start by drawing the diagram for a better visualisation of the situation given in the question.

It is given that AB=CD and BC=AD. Also, we can see that AC is common in $\mathrm{\Delta }ABC\text{ and }\mathrm{\Delta }CDA$ . Therefore, as all three sides of $\mathrm{\Delta }ABC$ are equal to corresponding sides of $\mathrm{\Delta }CDA$ , we can say that both the triangles are congruent by SSS congruence rule.
Therefore, by using CPCT, we can say that:
$\begin{array}{rl}& \mathrm{\angle }ABC=\mathrm{\angle }CDA\\ & \mathrm{\angle }CAB=\mathrm{\angle }ACD...........(i)\\ & \mathrm{\angle }DAC=\mathrm{\angle }BCA............(ii)\end{array}$
Now if we add equation (i) and equation (ii), we get
$\mathrm{\angle }CAB+\mathrm{\angle }DAC=\mathrm{\angle }ACD+\mathrm{\angle }BCA$
Now from the figure, we can deduce that $\mathrm{\angle }CAB+\mathrm{\angle }DAC=\mathrm{\angle }DAB$ and $\mathrm{\angle }ACD+\mathrm{\angle }BCA=\mathrm{\angle }BCD$. Therefore, our equation becomes:
$\mathrm{\angle }DAB=\mathrm{\angle }BCD$
As the opposite angles of quadrilateral ABCD are equal, i.e., $\mathrm{\angle }DAB=\mathrm{\angle }BCD$ and $\mathrm{\angle }ABC=\mathrm{\angle }CDA$ , so we can say that quadrilateral ABCD is a parallelogram.

Note: It is prescribed to learn all the basic theorems related to congruence and similarity of triangles as they are used quite often. Also, learn the properties of parallelograms, including squares, as they might also be needed for solving such problems as we used in the above question.