Question
ABCD is a rectangle in which diagonal BD bisects $\angle\text{B}.$ Show that ABCD is a square.
✓
Answer
Given: A rectangle ABCD in which diagonal BD bisects $\angle\text{B}$
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.
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