Question

Question 8
ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

Answer 1

In $\Delta ABDand\Delta ACD,$
AB = AC [given]
AD = AD [common side]
and BD = DC [given]
$\therefore \Delta ABD\cong \Delta ACD$ [by SSS congruence rule]
$\Rightarrow \angle BAD=\angle CAD$ [by CPCT]
i.e., $\angle BAO=\angle CAO$


In $\Delta AOBand\Delta AOC,$
AB = AC [given]
AO = OA [common side]
and $\angle BAO=\angle CAO$ [proved above]
$\therefore \Delta AOB\cong \Delta AOC$ [by SAS congruence rule]
$\Rightarrow BO=OC$ [by CPCT]
and $\angle AOB=\angle AOC$ [by CPCT]...(i)
But $\angle AOB+\angle AOC={180}^{\circ }$ [linear pair axiom]
$\Rightarrow \angle AOB+\angle AOB={180}^{\circ }$ [from Eq. (i)]
$\Rightarrow 2\angle AOB={180}^{\circ }\Rightarrow \angle AOB=\frac{{180}^{\circ }}{2}={90}^{\circ }$
Hence, AD $\perp$ BC and AD bisects BC i.e., AD is the perpendicular bisector of BC.