Answer 1

We have,

∠C = 90° and D is the mid-point of BC

In ΔACB, by Pythagoras theorem

AB² = AC² + BC²

⇒ AB² = AC² + (2CD)²                  [D is the mid-point of BC]

AB² = AC² + 4CD²

⇒ AB² = AC² + 4(AD² − AC²)       [In ΔACD, by Pythagoras theorem]

⇒ AB² = AC² + 4AD² − 4AC²

⇒ AB² = 4AD² − 3AC²