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In an equilateral triangle 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:
We have to prove that $9AD^{2}=7AB^{2}$.
Solution:
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]
$AB=AC=x$ [Equilateral triangle]
$\angle AEB=\angle AEC=90^o$ [$AE\perp BC$]
Therefore, by R.H.S. congruency,
$\vartriangle AEB \cong \vartriangle AEC$
This implies,
$BE=EC$ [C.P.C.T.]
$BE=EC=\frac{x}{2}$
$\Rightarrow BD+DE=\frac{x}{2}$
$\Rightarrow \frac{x}{3}+DE=\frac{x}{2}$
$\Rightarrow DE=\frac{x}{2}-\frac{x}{3}$
$\Rightarrow DE=\frac{x}{6}$
Using Pythagoras Theorem,
$( Hypotenuse)^{2}=( Height)^{2}+( Base)^{2}$
In $\vartriangle AEB$,
$(AB)^{2}=(AE)^{2}+(BE)^{2}$
$x^{2}=(AE)^{2}+( \frac{x}{2})^{2}$
$\Rightarrow (AE)^{2}=x^{2}-\frac{x^{2}}{4}$
$\Rightarrow ( AE)^{2}=\frac{4x^{2}-x^{2}}{4}$
$\Rightarrow ( AE)^2=\frac{3x^{2}}{4}$.........(i)
Similarly, In $\vartriangle AED$
$(AD)^{2}=(AE)^{2}+(DE)^{2}$
$\Rightarrow (AD)^{2}=\frac{3x^{2}}{4}+( \frac{x}{6})^{2}$ [From (i)]
$\Rightarrow (AD)^{2}=\frac{3x^{2}}{4}+\frac{x^{2}}{36}$
$\Rightarrow (AD)^{2}=\frac{27x^{2}+x^{2}}{36}$
$\Rightarrow (AD)^{2}=\frac{28x^{2}}{36}$
$\Rightarrow (AD)^{2}=\frac{7x^{2}}{9}$
$\Rightarrow 9(AD)^{2}=\frac{7x^{2}}{9}\times 9$ [Multiplying both sides by 9]
$\Rightarrow 9(AD)^{2}=7x^{2}$
$\Rightarrow 9(AD)^{2}=7(AB)^{2}$
Hence proved.