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
12