abcd is a rectangle in which …
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Sia ? 6 years, 6 months ago
Given: A rectangle ABCD in which diagonal BD bisects {tex}\angle B{/tex}.
To prove: ABCD is a square.
Proof: DC||AB [{tex}\because{/tex} Opposite sides of a rectangle are parallel]
{tex}\Rightarrow \;\angle 4 = \angle 1{/tex} ..(1) [Alternate interior angles]
Similarly, {tex}\angle 3 = \angle 2{/tex} …(2) [Alternate interior angles]
And {tex}\angle 1 = \angle 2{/tex}…(3) [Given]
From equation (1), (2) and (3), we get
{tex}\angle 3 = \angle 4{/tex}
In {tex}\Delta BDA{/tex} and {tex}\Delta BDC{/tex}, we have
{tex}\angle 1 = \angle 2{/tex} [Given]
BD = BD [Common side]
{tex}\angle 3 = \angle 4{/tex} [proved above]
So, By ASA criterion of congruence, we have
{tex}\Delta BDA \cong \Delta BDC{/tex}
{tex}\therefore{/tex} AB = BC [CPCT]
So, ABCD is a square.
Hence, proved.
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