Given: ΔABC is an isosceles triangle with AB = AC = 13 cm
Suppose the altitude from A on Bc meets BC at M.
∴ M is the midpoint of BC. AM = 5 cm
In ∆AMB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AM)2 + (BM)2 = (AB)2
⇒ (5)2 + (BM)2 = (13)2
⇒ (BM)2 = (13)2 – (5)2
⇒ (BM)2 = (13 – 5)(13+5)
[∵ (a2 – b2) = (a + b)(a – b)]
⇒ (BM)2 = (8)(18)
⇒ (BM)2 = 144
⇒ BM = ±12
⇒ BM = 12 [taking positive square root]
∴ BC = 2BM or 2MC = 2×12 = 24cm
