Question

In a right-angled triangle prove that 4 times the sum of the squares on the medians drawn from the acute angles is equal to 5 times the square on the hypotenuse.
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Answer 1

Figure is in the attachment...

In figure ∆ABC, hypotenuse = AC.

Medians = AD & CE.

So we have to prove ; 5AC² = 4(AD² + CE²)

Proof

In ∆ABC ; using Pythagoras theorem :

→ AC² = AB² + BC² ---------- (1)

Now as AD is median drawn on BC, so BD = CD = BC/2

And as CE is median drawn on AB, so AE = BE = AB/2

Now using ∆ABD, using Pythagoras theorem :

→ AD² = AB² + BD²

→ AD² = AB² + (BC/2)²

→ AD² = AB² + (BC²/4)

→ AD² = (4AB² + BC²)/4

→ 4AD² = 4AB² + BC² ------- (2)

Now using ∆BCE ; using Pythagoras theorem :

→ CE² = BE² + BC²

→ CE² = (AB/2)² + BC²

→ CE² = (AB²)/4 + BC²

→ CE² = (AB² + 4BC²)/4

→ 4CE² = 4BC² + AB² --------- (3)

Now adding (2) and (3) :

→ 4AD² + 4CE² = 4AB² + BC² + 4BC² + AB²

→ 4(AD² + CE²) = 5AB² + 5BC²

→ 4(AD² + CE²) = 5(AB² + BC²)

From (1) : AB² + BC² = AC²

→ 5AC² = 4(AD² + CE²) [Proved]

Hence proved ; in a right angled triangle 4 times the sum of the squares on the medians drawn from the acute angles is equal to 5 times the square on the hypotenuse.

Answer 2

Step-by-step explanation:

hope u would have understood ♡♡