Question

In a quadrilateral $ABCD$, $\angle$$A$=$\angle$$C$ and $\angle$$B$=$\angle$$D$. Then, $ABCD$ is a square.

• A. True
• B. False

Answer 1

The correct option is B False

If both the pairs of opposite angles of a quadrilateral are congruent then it is a parallelogram.
Proof - Since, the sum of the angles of quadrilateral is ${360}^{\circ }$
$\Rightarrow$ $\angle$$A$+ $\angle$$B$ + $\angle$$C$ + $\angle$$D$ = ${360}^{\circ }$
$\Rightarrow$ $\angle$$A$ + $\angle$$D$ + $\angle$$A$ + $\angle$$D$ =${360}^{\circ }$
$\Rightarrow$ 2$\angle$$A$= 2$\angle$$D$=${360}^{\circ }$
$\Rightarrow$ $\angle$$A$ + $\angle$$D$ = ${180}^{\circ }$ [Co-interior angle]
$\Rightarrow$ AB ║ DC Similarly, $\angle$$A$ + $\angle$$B$ + $\angle$$C$ + $\angle$$D$ = ${360}^{\circ }$
$\Rightarrow$ $\angle$$A$ + $\angle$$B$ + $\angle$$A$ + $\angle$$B$ = ${360}^{\circ }$ [∠A = ∠C and ∠B = ∠D]
$\Rightarrow$ 2 $\angle$$A$ + 2$\angle$$B$ = ${360}^{\circ }$
$\Rightarrow$ $\angle$$A$ + $\angle$$B$ = ${180}^{\circ }$ [∵This is sum of interior angles on the same side of transversal AB]
$\therefore$ AD ║ BC
So, AB ║ DC and AD ║ BC
$\Rightarrow$ ABCD is a parallelogram.