Question

If ABCD is an isosceles trapezium, then $\mathrm{\angle }C$ is equal to:
(A). $\mathrm{\angle }B$
(B). $\mathrm{\angle }A$
(C). $\mathrm{\angle }D$
(D). $90{}^{\circ }$

Answer 1

Hint: Start by drawing the diagram with all required constructions. Use the property that in isosceles trapezium the lengths of the non-parallel sides are equal.

Complete step-by-step solution -
Let us start with the solution by drawing a diagram of the situation given in the figure along with the required constructions for better visualisation.

In the above figure, we have constructed lines AE and DF such that they are perpendicular to both AD and BC.
Now, as it is given that ABCD is an isosceles trapezium, the lengths of the non-parallel sides must be equal, i.e., AB=DC.
Now, if we focus on $\mathrm{\Delta }ABE$ and $\mathrm{\Delta }DCF$ , we have
AB=DC
We also know that AE=DF, as the perpendicular distance between two parallel lines are equal at any point on the lines. And $\mathrm{\angle }AEB=\mathrm{\angle }DFC=90{}^{\circ }$ , as constructed by us, AE and DF are the heights of the trapezium.
Therefore, using the RHS rule of congruency, we can say that $\mathrm{\Delta }ABE\cong \mathrm{\Delta }DCF$ .
So, using the rule of CPCT, we can deduce that: $\mathrm{\angle }B=\mathrm{\angle }C$ .
Therefore, the answer to the above question is option (a).

Note: Some, other important properties of the isosceles trapezium are:
The diagonal of isosceles trapezium bisect each other and the sum of opposite angles is equal to $180{}^{\circ }$ . All the properties of a trapezium are also valid for isosceles trapezium.