Question

In given fig, ΔABC, < B is right angle and D is midpoint of BC. Prove that AC^2 = 4AD^2 - 3AB^2




please answer​

Answer 1

In right triangle ABC,

By Pythagorous theorem,

AC² = BC² + AB²     ----- (1)

Likewise,

In right angle triangle ABD,

AD² = BD² + AB²

       = (1/2 BC)² + AB²

      = 1/4 BC² + AB²

AD² = 1/4 BC² + AB²

=> BC² = 4AD² - 4AB²     ------ (2)

Solve (1) & (2)

=> AC² = 4AD² - 4AB² + AB²

=> AC² = 4AD² - 3AB²

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