Question

3.
P and Q are midpoints of side AC and BC respectively of triangle ABC right angled at C prove that 4BP² = 4 BC² + AC²
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Answer 1

Given :-

• P and Q are midpoints of sides AC and BC respectively
• Δ ABC is right angled at C

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To prove :-

• 4 BP² = 4 BC² + AC²

Knowledge required :-

• Pythagoras theorem

Pythagoras theorem states that In any right angled triangle The square of hypotenuse is equal to the sum of the squares of other two sides of triangle .

Proof :-

Taking RHS

RHS = 4 BC² + AC²

( using AC = 2 CP )

RHS = 4 BC² +  ( 2 CP )²

RHS = 4 BC² + 4 CP²

RHS = 4 ( BC² + CP² )

( ∵ Δ PCB is right angled triangle

∴By pythagoras theorem  BC² + CP²= BP²)

so,

RHS = 4 ( BP² )

RHS = 4 BP² = LHS

ʟʜs = ʀʜs

Hence, Proved .