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.
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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