Question

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

Answer 1

Given:

Equilateral triangle $ABC$

$D$ is a point on $BC$

$BD=\frac{1}{2}BC$

To prove:

$9A{D}^2=7A{B}^2$

Construction:

Let us draw $AE\perp BC$

Consider the diagram shown below.

Proof:

Here,

$AB=BC=AC=x\left(say\right)$

So,

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

In $\Delta AEB$ and $\Delta AEC$,

$AE=AE$ (Common)

$AB=AC$ (Both are $x$ as it is an equilateral triangle)

$\angle AEB=\angle AEC$

So, by RHS congruency,

$\Delta AEB\cong \Delta AEC$

So,

$\Rightarrow BE=EC$ (CPCT)

Now,

$BE=EC=\frac{1}{2}BC$

$\Rightarrow BE=EC=\frac{x}{2}$

Therefore,

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

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

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

In right angled $\Delta AEB$, using Pythagoras theorem, we have

$A{B}^2=A{E}^2+B{E}^2$

$x^2=A{E}^2+\frac{x^2}{4}$

$A{E}^2=\frac{3x^2}{4}$

Similarly, in $\Delta AED$,

$A{D}^2=A{E}^2+D{E}^2$

$A{D}^2=\frac{3x^2}{4}+\frac{x^2}{36}$

$A{D}^2=\frac{28x^2}{36}$

$A{D}^2=\frac{7x^2}{9}$

$9A{D}^2=7x^2$

$\Rightarrow 9A{D}^2=7A{B}^2$

Hence, proved.