Question

In A ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that A ABC is an isosceles triangle in which AB = AC.

Answer 1

Let the sign for congruence be '='.

In ∆ADB and ∆ADC,
seg AD = seg AD ...(Common side)
Angle ADB = Angle ADC ...(Both 90 degrees)
Angle BAD = Angle CAD ...(seg AD bisects angle BAC - Given)
Thus, ∆ADB = ∆ADC ...(i) (SAA Test of congruence of triangles)
Thus, from equation (i), we can conclude,
seg AB = seg AC ...(ii) (Congruent sides of congruent triangles)

Now, in ∆ABC,
side AB = side AC ...[From (ii)]
Thus, ∆ABC is an isosceles triangle in which AB = AC.

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer: -}}}}$

Given,

AD is the perpendicular bisector of BC

To show,

AB = AC

Proof,

In ΔADB and ΔADC,

AD = AD (Common)

∠ADB = ∠ADC

BD = CD (AD is the perpendicular bisector)

Therefore, ΔADB ≅ ΔADC by SAS congruence condition.

AB = AC (by CPCT) .