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.Need a full question paper?
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