Question
$ABCD$ is a quadrilateral such that diagonal $AC$ bisects the angles $A$ and $C$. Prove that $AB = AD$ and $CB = CD$.

Answer

Given in a quadrilateral $ABCD$, diagonal $AC$ bisects the angles $A$ and $C.$
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To prove $\text{AB}=\text{CD}\ \text{ and }\text{CB}=\text{CD}$ Proof in $\triangle\ \text{ADC}\ \text{and}\ \triangle\ \text{ABC},$
$\angle\ \text{DAC}=\angle\ \text{BAC}$
$\big[\because$ AC is the bisector of $\angle\ \text{A}\ \text{and}\ \angle\ \text{C}\big]$
$\angle\ \text{DCA}=\angle\ \text{BCA}$
$\big[\because$ AC is the bisector of $\angle\ \text{A}\ \text{and}\ \angle\text{C}\big]$
$\text{and}\ \text{AC}=\text{AC}$ [common side]
$\therefore\ \triangle\text{ADC}\cong\triangle\text{ABC}$ [by$ASA$ congruence rule]
$\text{AD}=\text{AB}$ [by CPCT] $\text{and}\ \text{CD}=\text{CB}$ [by $CPCT$] Hence proved.

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