Quadrilaterals — Maths STD 9 — Question

To prove: ABCD is a square.

Proof: DC || AB [Opposite sides of a rectangle are parallel]

$\Rightarrow\ \angle4=\angle1\ ...(\text{i})$ [Alternate interior angles]

Similarly, $\angle3=\angle2\ ...(\text{ii})$ [Alternate interior angles]

And $\angle1=\angle2\ ...(\text{iii})$ [Given]

From equation (1), (2) and (3), we get

$\angle3=\angle4$

In $\Delta\text{BAD}$ and $\Delta\text{BDC}$ we have

$\angle1=\angle2$[Given]

BD = BD[Common side)

$\angle3=\angle4$ [proved above]

So, By ASA criterion of congruence, we have

$\Delta\text{BAD}\cong\Delta\text{BCD}$

$\therefore\ \text{AB}=\text{BC}$ [CPCT]

As, adjacent sides of rectangle are equal. So, ABCD is a square.