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.

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Solution

Let $A B C$ be a right angled triangle $A B = x$ , $B C = y$ $∴$ $A C = \sqrt{x^{2} + y^{2}}$

Let $A D$ be medium on $B C$ & $C E$ be medium on $A B$

$A D = \sqrt{x^{2} + \frac{y^{2}}{4}}$

$C E = \sqrt{y^{2} + \frac{x^{2}}{4}}$

${A D}^{2} + {C E}^{2} = x^{2} + \frac{x^{2}}{4} + y^{2} + \frac{y^{2}}{4} = \frac{5}{4} (x^{2} + y^{2})$

$\Rightarrow 4 ({A D}^{2} + {C E}^{2}) = 5 (x^{2} + y^{2})$

$\Rightarrow 4 ({A D}^{2} + {C E}^{2}) = 5 ({A C}^{2})$

Hence Proved.

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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.

Let ABC be a right angled triangle AB=x, BC=y ∴ AC=√x2+y2Let AD be medium on BC & CE be medium on ABAD=√x2+y24CE=√y2+x24AD2+CE2=x2+x24+y2+y24=54(x2+y2)⇒4(AD2+CE2)=5(x2+y2)⇒4(AD2+CE2)=5(AC2)Hence Proved.