In an equilateral $\vartriangle ABC$, $D$ is a point on side $BC$ such that $BD=\frac{1}{3}BC$. Prove that $9AD^{2}=7AB^{2}$.

Given: Equilateral $\vartriangle ABC$, D is a such a point on side BC that $BD=\frac{1}{3}BC$.

To do: To Prove that $9AD^{2}=7AB^{2}$.

Solution:  

Let us draw $AE\perp BC$.

All sides of equilateral; triangles are equal ,

$\therefore AB=BC=CA$

Let $AB=BC=CA=x$

As given 

$BD=\frac{1}{3}BC$

$\Rightarrow BD=\frac{x}{3}$

In $\vartriangle AEB$  and $vartriangle AEC$,

$AE=AE$                      [common]

$AB=AC$                      [Both x as it is an equilateral triangle]

$\angle AEB=\angle AEC$    [Both $90^{o}$ as $AE\perp BC$]

Hence By R.H.S. congruency,

$\vartriangle AEB \cong \vartriangle AEC$

$\therefore BE=EC$           [By C.P.C.T.]

So, $BE=EC=\frac{x}{2}$

$\because BE=\frac{x}{2}$

$\therefore BD+DE=\frac{x}{2}$

$\Rightarrow \frac{x}{3}+DE=\frac{x}{2}$

$\Rightarrow DE=\frac{x}{2}-\frac{x}{3}$

$\Rightarrow DE=\frac{x}{6}$

Using Pythagoras Theorem,

$( Hypotenuse)^{2}=( Height)^{2}+( Base)^{2}$

In $\vartriangle AEB$,

$(AB)^{2}=(AE)^{2}+(BE)^{2}$

$x^{2}=(AE)^{2}+( \frac{x}{2})^{2}$

$\Rightarrow (AE)^{2}=x^{2}-\frac{x^{2}}{4}$

$\Rightarrow ( AE)^{2}=\frac{4x^{2}-x^{2}}{4}$

$\Rightarrow ( AE)^2=\frac{3x^{2}}{4}$          ......$( 1)$

Similarly, In $\vartriangle AED$

$(AD)^{2}=(AE)^{2}+(DE)^{2}$

$\Rightarrow (AD)^{2}=\frac{3x^{2}}{4}+( \frac{x}{6})^{2}$   From $( 1)$

$\Rightarrow (AD)^{2}=\frac{3x^{2}}{4}+\frac{x^{2}}{36}$

$\Rightarrow (AD)^{2}=\frac{27x^{2}+x^{2}}{36}$

$\Rightarrow (AD)^{2}=\frac{28x^{2}}{36}$

$\Rightarrow (AD)^{2}=\frac{7x^{2}}{9}$

$\Rightarrow 9(AD)^{2}=\frac{7x^{2}}{9}\times 9$       [On multiplying both sideas by 9]

$\Rightarrow 9(AD)^{2}=7x^{2}$

$\Rightarrow 9(AD)^{2}=7(AB)^{2}$

Hence Proved.

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Updated on: 10-Oct-2022

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