Question
$ABCD$ is quadrilateral such that $AB = AD$ and $CB = CD.$ Prove that $AC$ is the perpendicular bisector of $BD.$
✓
Answer
Given In quadrilateral $ABCD, AB = AD$ and $CB = CD$. constrion: To prove $AC$ is the petpendicular bisector of $BD.$ uction Join $AC$ and $BD.$
Proof In $\triangle\text{ABC}\text{ and }\triangle\text{ADC},$
$\text{AB}=\text{AD}$ [given] $\text{BC}=\text{CD}$ [given] $\text{and}\ \text{AC}=\text{AC}$ [common side]
$\therefore \triangle\text{ABC}\cong\triangle\text{ADC}[$ by $SSS$ congruence rule$]$
$\Rightarrow\ \angle1=\angle2[$ by $CPCT]$
now, in $\triangle\text{AOB}\text{ and }\triangle\text{AOD},\ \text{AB}=\text{AD}$ [given]
$\Rightarrow\ \angle1=\angle2[ $proved above$]$
$\text{and}\ \text{AO}=\text{AO} [$common side$]$
$\therefore\ \triangle\text{AOB}\cong\triangle\text{AOD}[$ by $SAS$ congruence rule$]$
$\Rightarrow\ \text{BO}=\text{DO} [$by $CPCT]$
$\text{and}\ \angle3=\angle4 [$by $CPCT]...(i) $
But $\angle3+\angle4=180^\circ$[linear pair axiom]
$\angle3+\angle3=180^\circ [$from Eq. $(i)]$
$\Rightarrow\ 2\angle3=180^\circ$
$\Rightarrow\ \angle3=\frac{180^\circ}{2}$
$\therefore\ \angle3=90^\circ$ i.e., $AC$ is perpendicular bisector of $BD.$
Proof In $\triangle\text{ABC}\text{ and }\triangle\text{ADC},$
$\text{AB}=\text{AD}$ [given] $\text{BC}=\text{CD}$ [given] $\text{and}\ \text{AC}=\text{AC}$ [common side]
$\therefore \triangle\text{ABC}\cong\triangle\text{ADC}[$ by $SSS$ congruence rule$]$
$\Rightarrow\ \angle1=\angle2[$ by $CPCT]$
now, in $\triangle\text{AOB}\text{ and }\triangle\text{AOD},\ \text{AB}=\text{AD}$ [given]
$\Rightarrow\ \angle1=\angle2[ $proved above$]$
$\text{and}\ \text{AO}=\text{AO} [$common side$]$
$\therefore\ \triangle\text{AOB}\cong\triangle\text{AOD}[$ by $SAS$ congruence rule$]$
$\Rightarrow\ \text{BO}=\text{DO} [$by $CPCT]$
$\text{and}\ \angle3=\angle4 [$by $CPCT]...(i) $
But $\angle3+\angle4=180^\circ$[linear pair axiom]
$\angle3+\angle3=180^\circ [$from Eq. $(i)]$
$\Rightarrow\ 2\angle3=180^\circ$
$\Rightarrow\ \angle3=\frac{180^\circ}{2}$
$\therefore\ \angle3=90^\circ$ i.e., $AC$ is perpendicular bisector of $BD.$
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