Question

Question 21
If ABCD is quadrilateral such that AB = AD and CB = CD then prove that AC is the perpendicular bisector of BD.

Answer 1

Given: In quadrilateral ABCD, AB = AD and CB = CD.
Construction: Join AC and BD.
To prove: AC is the perpendicular bisector of BD.

Proof: In $\Delta ABCand\Delta ADC,$
AB = AD [given]
BC = CD [given]
and AC = AC [common side]
$\therefore \Delta ABC\cong \Delta ADC$ [by SSS congruence rule]
$\Rightarrow \angle 1=\angle 2$ [by CPCT]
Now, in $\Delta AOBand\Delta AOD,$ AB = AD [given]
$\Rightarrow \angle 1=\angle 2$ [proved above]
and AO = AO [common side]
$\therefore \Delta AOB\cong \Delta AOD$ [by SAS congruence rule]
$\Rightarrow BO=DO$ [by CPCT]
and $\angle 3=\angle 4$ [by CPCT]...(i)
But $\angle 3+\angle 4={180}^{\circ }$ [linear pair axiom]
$\angle 3+\angle 3={180}^{\circ }$ [from Eq. (i)]
$\Rightarrow 2\angle 3={180}^{\circ }$
$\Rightarrow \angle 3=\frac{{180}^{\circ }}{2}\therefore \angle 3={90}^{\circ }$
i.e., AC is perpendicular bisector of BD.