ABCD is a rectangle in which diagonal AC bisects as well as . Show that: i. ABCD is a square ii. diagonal BD bisects as well as .

i. ABCD is a rectangle in which diagonal AC bisects as well as . = ...(1) And = ...(2) Since every rectangle is a parallelogram, therefore AB | DC and AC is the transversal. = (alternate angles) = [from] Thus in , AD = CD (opposite sides of equal angles are equal) But, AD = BC and CD = AB (ABCD is a rectangle) = BC = CD = AD Hence, ABCD is a square. ii. In and , AB = CD AD = BC BD = BD (by SSS congruence criterion) = and = [CP.C.T.] Hence, diagonal BD bisects as well as .

ABCD is a rectangle in which diagonal AC bisects as well as . Sho…

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Question

$\text{ABCD}$ is a rectangle in which diagonal $AC$ bisects $\angle\text{A}$ as well as $\angle\text{C}.$ Show that:
$i. \text{ABCD}$ is a square
$ii.$ diagonal $BD$ bisects $\angle\text{B}$ as well as $\angle\text{D}.$

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Answer

$i. \text{ABCD}$ is a rectangle in which diagonal $AC $ bisects $\angle\text{A}$ as well as $\angle\text{C}.$
$\Rightarrow\angle\text{BAC}=\angle\text{DAC}...(1)$
And $\angle\text{BCA}=\angle\text{DCA}...(2)$
Since every rectangle is a parallelogram, therefore
$AB\ \|\ DC$ and $AC $ is the transversal.
$\Rightarrow\angle\text{BAC}=\angle\text{DCA}($alternate angles$)$
$\Rightarrow\angle\text{DAC}=\angle\text{DCA} [$from$]$
Thus in $\triangle\text{ADC},$
$\text{AD = CD} ($opposite sides of equal angles are equal$)$
But, $\text{AD = BC}$ and $\text{CD = AB} (\text{ABCD}$ is a rectangle$)$
$\Rightarrow\text{AB = BC = CD = AD}$
Hence, $\text{ABCD}$ is a square.
$ii.$ In $\triangle\text{BAD}$ and $\triangle\text{BCD},$
$\text{AB = CD}$
$\text{AD = BC}$
$\text{BD = BD}$
$\therefore\triangle\text{BAD}\cong\triangle\text{BCD} ($by $\text{SSS}$ congruence criterion$)$
$\Rightarrow\angle\text{ABD}=\angle\text{CBD}$ and $\angle\text{ADB}=\angle\text{CDB}[\text{CP.C.T.}]$
Hence, diagonal $BD$ bisects $\angle\text{B}$ as well as $\angle\text{D}.$