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Square

In a triangle...

Question

In a triangle ABC, if angle B = $90^{o}$ and D is the point in BC such that BC=2DC, then

{A C}^{2} = {A D}^{2} + 3 {C D}^{2}

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{A C}^{2} = {A D}^{2} + {5 C D}^{2}

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{A C}^{2} = {A D}^{2} + {7 C D}^{2}

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{A C}^{2} = {A B}^{2} + {5 B D}^{2}

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Solution

The correct option is B ${A C}^{2} = {A D}^{2} + {5 C D}^{2}$
Consider the figure shown. We need to a construction by extending BC to BD such that BC = 2 CD.
From $Δ A B D, {A D}^{2} = {A B}^{2} + B D^{2} [b y P y t h a g o r a s t h e o r e m] ⟹ {A D}^{2} = {A B}^{2} + (3 x)^{2} ⟹ {A D}^{2} = {A B}^{2} + 9 x^{2}$
From $Δ A B C, {A C}^{2} = {A B}^{2} + B C^{2} [b y P y t h a g o r a s t h e o r e m] ⟹ {A C}^{2} = {A B}^{2} + (2 x)^{2} ⟹ {A D}^{2} = {A B}^{2} + 4 x^{2}$
Now from $Δ A B D, {A D}^{2} = {A B}^{2} + B D^{2} ⟹ {A D}^{2} = {A B}^{2} + 9 x^{2} S o, {A B}^{2} = {A D}^{2} - 9 x^{2}$ .
Substituting for ${A B}^{2} i n {A C}^{2}, w e g e t {A C}^{2} = {A D}^{2} - 9 x^{2} + 4 x^{2} {A C}^{2} = {A D}^{2} - 5 x^{2} ⟹ A C^{2} + 5 x^{2} = {A D}^{2} ⟹ A C^{2} + 5 {C D}^{2} = {A D}^{2}$ .

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Similar questions

A B C, ∠ B = 90^{0}

D

B C

A C^{2} = A D^{2} + 3 C D^{2}

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i)

(ii)

(iii)

Δ

∠ B = 90^{0}

A C^{2} = A D^{2} + 3 C D^{2}

∠ B = 90^{\circ}

{A C}^{2} = {A D}^{2} + 3 {C D}^{2}

A B C, ∠ B = 90^{\circ}

D

A C^{2} = A D^{2} + 3 C D^{2}

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