In a right angled triangle, five times the square on the hypotenuse is equal to four times the sum of the squares on the medians drawn from the acute angles. Prove it.

Given: ABC is a right-angled triangle at B and AD and CE are the medians drawn from the acute angles.

To prove: 5AC² = 4(AD² + CE²)

Proof: Let the lengths of AB and BC be x and y, respectively.

In Δ ABC, using Pythagoras theorem,

AC² = AB² + BC²

⇒ AC² = x² + y²

⇒ AC = $\sqrt{x^2 + y^2}$

In Δ ABD, using Pythagoras theorem,

AD² = AB² + BD²

⇒ AD² = x² + $\Big(\dfrac{y}{2}\Big)^2$

⇒ AD² = x² + $\dfrac{y^2}{4}$ ……………(1)

In Δ BCE, using Pythagoras theorem,

EC² = EB² + BC²

⇒ EC² = $\Big(\dfrac{x}{2}\Big)^2$ + y²

⇒ EC² = $\dfrac{x^2}{4}$ + y² ……………(2)

Adding equation (1) and (2), we get

⇒ EC² + AD² = $\dfrac{x^2}{4}$ + y² + x² + $\dfrac{y^2}{4}$

⇒ EC² + AD² = $\dfrac{x^2}{4} + \dfrac{4y^2}{4} + \dfrac{4x^2}{4} + \dfrac{y^2}{4}$

⇒ EC² + AD² = $\dfrac{x^2}{4} + \dfrac{4x^2}{4} + \dfrac{y^2}{4} + \dfrac{4y^2}{4}$

⇒ EC² + AD² = $\dfrac{5x^2}{4} + \dfrac{5y^2}{4}$

⇒ EC² + AD² = 5 ($\dfrac{x^2 + y^2}{4}$)

⇒ 4(EC² + AD²) = 5(x² + y²)

⇒ 4(EC² + AD²) = 5AC²

Hence, five times the square on the hypotenuse is equal to four times the sum of the squares on the medians drawn from the acute angles.