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The Property of the Centroid of a Triangle

In the figure...

Question

In the figure, $A D$ is a median of a triangle $A B C$ and $A M ⊥ B C$ .
$2 A D^{2} + \frac{1}{2} B C^{2} =$

A C^{2} + B C^{2}

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A B^{2} + B C^{2}

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A C^{2} + A B^{2}

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none of the above

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Solution

The correct option is B $A C^{2} + A B^{2}$
From the triangle,
$A D$ is the median of the triangles
$\Rightarrow C D = B D = \frac{B C}{2}$ ...(i)
By Pythagoras Theorem, $△ A M C$
${A C}^{2} = {C M}^{2} + {A M}^{2}$ ...(ii)
In $△ A B M,$ ${A B}^{2} = {B M}^{2} + {A M}^{2}$ ...(iii)
In $△ A M D,$ ${A D}^{2} = {D M}^{2} + {A M}^{2}$ ...(iii)
$\Rightarrow {A M}^{2} = {A D}^{2} - {D M}^{2}$ ...(iv)
$\Rightarrow$ Substitute (iv) in (ii)
${A C}^{2} = {C M}^{2} + {A D}^{2} - {D M}^{2}$
$∴ {A C}^{2} = {(C D + D M)}^{2} + {A D}^{2} - {D M}^{2}$
${A C}^{2} = {(\frac{B C}{2} + D M)}^{2} + {A D}^{2} - {D M}^{2}$
$= \frac{{B C}^{2}}{2} + B C \times D M + {A D}^{2} + {D M}^{2} - D M$
${A C}^{2} = \frac{{B C}^{2}}{2} + B C \times D M + {A D}^{2}$ ....(v)
$\Rightarrow$ Substitute (iv) in (iii)
${A B}^{2} = {B M}^{2} + {A D}^{2} - {D M}^{2}$
$\Rightarrow {A B}^{2} = {(B D - D M)}^{2} + {A D}^{2} - {D M}^{2}$
$= (\frac{B C}{2} - D M) + {A D}^{2} - {D M}^{2}$
$\frac{{B C}^{2}}{4} - B C \times D M + {A D}^{2} + {D M}^{2} - {D M}^{2}$
$\Rightarrow {A B}^{2} = \frac{{B C}^{2}}{4} - B C \times D M + {A D}^{2}$ ...(vi)
Adding (v) and (vi)
${A B}^{2} + {A C}^{2} = \frac{{B C}^{2}}{4} + {A D}^{2} - B C \times D M + \frac{{B C}^{2}}{4} + B C \times D M + {A D}^{2}$
$\Rightarrow {A B}^{2} + {A C}^{2} = \frac{{B C}^{2}}{2} + 2 {A D}^{2}$

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Similar questions

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

(i)

(ii)

(iii)

Δ A B C

A D

A M ⊥ B C

{A B}^{2} + {A C}^{2} = 2 {A D}^{2} + \frac{1}{2} {B C}^{2}

△ A B C, A D

A E ⊥ B C

A B^{2} + A C^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}

△ A B C

⊥

A C^{2} = A D^{2} + B C \times D M + \frac{B C^{2}}{4}

A B^{2} = A D^{2} - B C \times D M + \frac{B C^{2}}{4}

A B C

∠ A B C < 90^{\circ}

A D ⊥ B C .

A C^{2} = A B^{2} + B C^{2} - 2 B C . D B .

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