Question

ABC and DBC are two isosceles triangle on the same base BC such that A and D lies on the opposite sides of BC. Show that AD is the perpendicular bisector of BC.

Answer 1

Let $AD$ intersect $BC$ at $O$

Then we have to prove that $\angle AOB=\angle AOC={90}^{\circ }$ and $BO=OC$

In $△ABD$ and $△ACD$ we have

$AB=AC$(given)

$AD=DA$(common)

$BD=DC$(given)

$\therefore △ABD\cong △ACD$

$\Rightarrow \angle 1=\angle 2$ proved above

$\therefore \angle AOB=\angle AOC$

But $\angle AOB+\angle AOC={180}^{\circ }$ (linear pair)

$\Rightarrow \angle AOB+\angle AOB={180}^{\circ }$

$\Rightarrow 2\angle AOB={180}^{\circ }$

$\therefore \angle AOB={90}^{\circ }$

Hence $AD$ is perpendicular to $BC$ and $AD$ bisects $BC$

$\therefore AD$ is the perpendicular bisector of $BC$