Question

AQ and BP are medians of a triangle ABC right angled at C. Prove that
4(AQ’ + BP) = 5 AB-​

Answer 1

Given :- AQ and BP are medians of a triangle ABC right angled at C. Prove that 4(AQ² + BP²) = 5AB² .

Answer :-

from image, in right angled ΔACQ, we have,

→ AQ² = AC² + CQ² { By pythagoras theorem. }

→ AQ² = AC² + (CB/2)²

→ AQ² = AC² + (CB²/4) ----------- Eqn.(1)

Similarly, in right angled ΔPCB, we have,

→ BP² = CB² + PC²

→ BP² = CB² + (AC/2)²

→ BP² = CB² + (AC²/4) ----------- Eqn.(2)

also, in right angled ∆ACB ,

→ AB² = AC² + CB² ------------ Eqn.(3)

Adding Eqn.(1) and Eqn.(2) , we get,

→ AQ² + BP² = AC² + CB² + (AC²/4) + (CB²/4)

→ AQ² + BP² = (AC² + AC²/4) + (CB² + CB²/4)

→ AQ² + BP² = (5AC² + 5CB²)/4

→ 4(AQ² + BP²) = 5(AC² + CB²)

Putting value of Eqn.(3) in RHS,

→ 4(AQ² + BP²) = 5AB² (Proved).

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