In △ABC, ∠B = 90° and D is the mid-point of BC. Prove that

(i) AC² = AD² + 3 CD²

(ii) BC² = 4 (AD² - AB²)

(i) Given,

ABC is a triangle such that ∠ABC = 90°. D is the mid-point of BC.

In △ ABD, using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AD² = AB² + BD² ……(1)

Similarly, in △ ABC,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (BD + DC)²

⇒ AC² = AB² + BD² + DC² + 2 × BD × DC

As D is the midpoint of BC, BD = DC.

⇒ AC² = AB² + BD² + CD² + 2 × CD × CD

⇒ AC² = AB² + BD² + CD² + 2CD²

⇒ AC² = AB² + BD² + 3CD²

Using equation (1), we get

⇒ AC² = AD² + 3CD²

Hence, proved that AC² = AD² + 3 CD².

(ii) In △ ABD,

⇒ AD² = AB² + BD²

BD = $\dfrac{\text{BC}}{2}$

$\Rightarrow \text{AD}^2 = \text{AB}^2 + \Big(\dfrac{\text{BC}}{2}\Big)^2 \\[1em] \Rightarrow \text{AD}^2 = \text{AB}^2 + \dfrac{\text{BC}^2}{4} \\[1em] \Rightarrow \text{AD}^2 - \text{AB}^2 = \dfrac{\text{BC}^2}{4} \\[1em] \Rightarrow \text{BC}^2 = 4 (\text{AD}^2 - \text{AB}^2)$

Hence, proved that BC² = 4 (AD² - AB²).