In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :

(i) 4AD² = 4AC² + BC²

(ii) 4BE² = 4BC² + AC²

(iii) 4(AD² + BE²) = 5AB²

(i) Since, D is mid-point of BC,

∴ CD = $\dfrac{BC}{2}$

In right triangle ACD,

By pythagoras theorem,

⇒ AD² = AC² + CD²

⇒ AD² = AC² + $\Big(\dfrac{BC}{2}\Big)^2$

⇒ AD² = AC² + $\dfrac{BC^2}{4}$

⇒ AD² = $\dfrac{4\text{AC}^2 + \text{BC}^2}{4}$

⇒ 4AD² = 4AC² + BC² …..(1)

Hence, proved that 4AD² = 4AC² + BC².

(ii) As E is mid-point of AC,

∴ CE = $\dfrac{AC}{2}$

⇒ AC = 2CE

BCE is right triangle,

By pythagoras theorem,

⇒ BE² = BC² + CE²

Multiplying both sides by 4 we get,

⇒ 4BE² = 4BC² + 4CE²

⇒ 4BE² = 4BC² + (2CE)²

⇒ 4BE² = 4BC² + AC² ……(2)

Hence, proved that 4BE² = 4BC² + AC².

(iii) As, ABC is a right triangle,

By pythagoras theorem we get,

⇒ AB² = AC² + BC²

Adding 1 and 2 from above parts we get,

⇒ 4AD² + 4BE² = 4AC² + BC² + 4BC² + AC²

⇒ 4(AD² + BE²) = 5AC² + 5BC²

⇒ 4(AD² + BE²) = 5(AC² + BC²)

⇒ 4(AD² + BE²) = 5AB².

Hence, proved that 4(AD² + BE²) = 5AB².