In figure given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB² = 4AD² - 3AC².

Since, D is mid-point of BC so we get,

DC = $\dfrac{BC}{2}$.

ABC is a right triangle.

By pythagoras theorem,

⇒ AB² = AC² + BC² …….(i)

ADC is a right triangle.

By pythagoras theorem,

⇒ AD² = AC² + DC² …….(ii)

⇒ AC² = AD² - DC²

⇒ AC² = AD² - $\Big(\dfrac{1}{2}\text{BC}\Big)^2$

⇒ AC² = AD² - $\dfrac{1}{4}\text{BC}^2$

⇒ AC² = $\dfrac{4\text{AD}^2 - \text{BC}^2}{4}$

⇒ 4AC² = 4AD² - BC²

⇒ AC² + 3AC² = 4AD² - BC²

⇒ AC² + BC² = 4AD² - 3AC²

⇒ AB² = 4AD² - 3AC² [….From (i)]

Hence, proved that AB² = 4AD² - 3AC².