In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

In ADB

AD²+BD²=AB²

5²+BD²=13²

25+BD²=169

BD²=169-25

BD²=144

BD=

BD=12cm

In ADB and In ADC

∠ADB=∠ADC =90

AB=AC=13cm

AD=AD (Common)

ADB≅ADC (By RHS condition)

BD=CD=12cm (c.p.c.t)

BC=BD+DC

BC=12+12

BC=24cm