In $Δ A B C, A D ⊥ B C$ where D lies on BC. Also $B D = 3 C D$ . then $2 A B^{2} = 2 A C^{2} + 2 B C^{2}$ .

True

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False

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Solution

The correct option is B False
Let the length of $C D = x$

then $B D = 3 C D = 3 x$
also $B C = B D + C D = 3 x + x = 4 x$

In $Δ A B C$
we have
$A B^{2} = B C \times B D$
$⟹ A B^{2} = 4 x \times 3 x$
$⟹ 2 A B^{2} = 24 x^{2} - (1)$

Also $A C^{2} = B C \times C D$
$⟹ A C^{2} = 4 x \times x ⟹ A C^{2} = 4 x^{2} ⟹ 2 A C^{2} = 8 x^{2} - (2)$

Subtracting eqs.(2) from eqs.(1)
$2 A B^{2} - 2 A C^{2} = 24 x^{2} - 8 x^{2} ⟹ = 16 x^{2} = (4 x)^{2} = B C^{2}$

Hence $2 A B^{2} = 2 A C^{2} + B C^{2}$
so Given statement is false.

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A D ⊥ B C

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△

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