Question
ABCD is a rectangle in which diagonal AC bisects $\angle$A as well as $\angle$C. Show that:
- ABCD is a square
- diagonal BD bisects $\angle$B as well as $\angle$D.
✓
Answer
Given : ABCD is a rectangle,
$\angle$A = $\angle$C
$\frac{1}{2} \angle A=\frac{1}{2} \angle C$
To prove: ABCD is a square proof:
- $\angle$DAC = $\angle$DCA (AC bisects A and C)
CD = DA (Sides opposite to equal angles are also equal)
However,
DA = BC and AB = CD (Opposite sides of a rectangle are equal)
AB = BC = CD = DA
ABCD is a rectangle and all of its sides are equal.
Hence, ABCD is a square - Let us join BD
In $\triangle$BCD,
BC = CD (Sides of a square are equal to each other)
$\angle$CDB = $\angle$CBD (Angles opposite to equal sides are equal)
However,
$\angle$CDB = $\angle$ABD (Alternating interior angles for AB $\parallel$ CD)
$\angle$CBD = $\angle$ABD
BD bisects $\angle$B
Also,
$\angle$CDB = $\angle$ABD
BD bisects $\angle$D.
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