Question

In $\Delta ABC,AD$ is the perpendicular bisects of BC sent Fig. show that $\Delta ABC$ is an isosceles triangle in which AB = AC.

Answer 1

Since AD is the ${\perp }_r$bisector

BD$=$DC …………….$\left(1\right)$

In right $\Delta$ABD,

using Pythagoras theorem

$A{B}^2=A{D}^2+B{D}^2$

$A{B}^2-A{D}^2=B{D}^2$ ……………$\left(2\right)$

In right $\Delta$ADC.

$A{C}^2=A{D}^2+C{D}^2$

$A{C}^2-A{D}^2=D{C}^2$ …………..$\left(3\right)$

From $\left(1\right)$ we have $B{D}^2=D{C}^2$

$\therefore A{B}^2-A{D}^2=A{C}^2-A{D}^2$ [from $\left(2\right)$ & $\left(3\right)$]

$\therefore A{B}^2=A{C}^2$

$AB=AC$

Hence proved.