Question

In the adjoining figure, D and E are mid points of the side BC and CA respectively of a ABC, right angled at C. Prove that: (i) 4BE2=4BC2+AC2 (ii) 4(AD2+BE2)=5AB2​

Answer 1

Answer:

Given :-

• A right angle triangle ABC right angled at C.
• D is the midpoint of BC
• E is the midpoint of CA.

To Prove :-

• (i) 4BE² = 4BC² + AC²
• (ii) 4(AD² + BE²) = 5AB²

Proof :-

D is the midpoint of BC ⇛ CD = DB = 1/2 BC........(1)

E is the midpoint of CA ⇛ AE = EC = 1/2 AC.........(2)

In right ∆ BEC

By pythagoras theorem,

⇛ BE² = EC² + BC²

⇛ BE² = (1/2 AC)² + BC² (Using (1))

⇛ BE² = 1/4 AC² + BC²

⇛ 4BE² = AC² + 4BC².........(3)

This, proves the (i) part.

Again, In right ∆ ACD

By pythagoras theorem,

⇛ AD² = DC² + AC²

⇛ AD² = (1/2 BC)² + AC² (Using (2))

⇛ AD² = 1/4 BC² + AC²

⇛ 4AD² = BC² + 4AC².........(4)

In right ∆ ABC

By pythagoras theorem,

⇛ AB² = BC² + AC² .........(5)

Adding (3) and (4), we get

⇛ 4BE² + 4AD² = AC² + 4BC² + BC² + 4AC²

⇛ 4(AD² + BE²) = 5BC² + 5AC²

⇛ 4(AD² + BE²) = 5AB²

$\large{\boxed{\boxed{\bf{Hence, \: Proved}}}}$

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