Question

In the given figure, triangle ABC is a right angle triangle with ∠B = 90° and D is mid point of side BC.
Prove that :
AC² = AD² + 3CD²

Answer 1

D is the midpoint of BC= AD = CD.

Angle B is a right angled triangle.

Consider ΔABC

AC^2 = AB^2 + BC^2 [Pythagoras theorem]

⇒ AC2 = AB^2 + (2BD)^2  (BC=BD+DC,BD=DC,So we can replace BC by 2BD.

⇒ AC^2 = AB^2 + 4BD^2 ----------- (1)

Consider ΔABC

AD^2 = AB^2 + BD^2 [Pythagoras theorem] ----------- (2)

Subtracting equation (2) from (1), we get

⇒ AC^2 - AD^2 = 3BD^2

⇒ AC^2 - AD^2 = 3CD^2 [  BD = CD,So we can replace BD by CD]

⇒ AC^2 = AD^2 + 3CD^2