Question

In an equilateral triangle $ABC$,$D$ is a point on side $BC$such that $BD=\frac{1}{3BC}$, prove that $9A{D}^2=7A{B}^2$.

Answer 1

Proof the given expression

Given: In an equilateral triangle $ABC$,$D$ is a point on side $BC$such that $BD=\frac{1}{3BC}$.

To prove: $9{\left(\text{AD}\right)}^2=7{\left(\text{AB}\right)}^2$

Let us draw an equilateral $△\text{ABC}$ with sides AB, BC and AC equal to each other. Also, draw a line from the vertex A which will meet the side BC at point D such that $\text{BD}={\displaystyle \frac{1}{3}}\left(\text{BC}\right)$. Also, draw a perpendicular from the vertex A on the side BC which divides line BC into two equal parts i.e., $\text{BE}=\text{CE}={\displaystyle \frac{\text{BC}}{2}}$

As, $\text{BD}=\frac{1}{3}\left(\text{BC}\right)....\left(1\right)$
Here, $\text{AB}=\text{BC}=\text{AC}$ and $\text{DE}=\left(\text{BE}-\text{BD}\right)$
Using Pythagoras theorem, In right angled $△\text{ADE}$,

$$\begin{gathered} {\left(\text{AD}\right)}^2={\left(\text{AE}\right)}^2+{\left(\text{DE}\right)}^2 \\ \Rightarrow {\left(\text{AD}\right)}^2={\left(\text{AE}\right)}^2+{\left(\text{BE}-\text{BD}\right)}^2.......\left(2\right) \end{gathered}$$

In right angled $△\text{ABE}$,
Using Pythagoras theorem

$$\begin{gathered} {\left(AB\right)}^2={\left(AE\right)}^2+{\left(BE\right)}^2 \\ \Rightarrow {\left(AE\right)}^2={\left(AB\right)}^2-{\left(BE\right)}^2 \\ \Rightarrow {\left(AE\right)}^2={\left(AB\right)}^2-{\left(\frac{BC}{2}\right)}^2......\left(3\right) \end{gathered}$$

Now using equation (3) in equation (2), we get
${\left(\mathrm{AD}\right)}^2={\left(\text{AB}\right)}^2-{\left(\frac{\text{BC}}{2}\right)}^2+{\left(\text{BE}-\text{BD}\right)}^2$
$\text{BE}={\displaystyle \frac{\text{BC}}{2}}$ and $\text{BD}={\displaystyle \frac{1}{3}}\left(\text{BC}\right)$ , so the above equation becomes
$$\begin{gathered} {\left(\mathrm{AD}\right)}^2={\left(\text{AB}\right)}^2-{\left(\frac{\text{BC}}{2}\right)}^2+{\left(\frac{\text{BC}}{2}-\frac{\text{BC}}{3}\right)}^2 \\ ={\left(\text{AB}\right)}^2-\left[\frac{{\left(\text{BC}\right)}^2}{4}\right]+{\left(\frac{\text{3BC}-\text{2BC}}{6}\right)}^2 \\ ={\left(\text{AB}\right)}^2-\left[\frac{{\left(\text{BC}\right)}^2}{4}\right]+\left[\frac{{\left(\text{BC}\right)}^2}{36}\right] \\ ={\left(\text{AB}\right)}^2+\left[\frac{{\left(\text{BC}\right)}^2}{36}-\frac{{\left(\text{BC}\right)}^2}{4}\right] \\ ={\left(\text{AB}\right)}^2+\left[\frac{{\left(\text{BC}\right)}^2-9{\left(\text{BC}\right)}^2}{36}\right] \\ ={\left(\text{AB}\right)}^2+\left[\frac{-8{\left(\text{BC}\right)}^2}{36}\right] \\ ={\left(\text{AB}\right)}^2-\left[\frac{2{\left(\text{BC}\right)}^2}{9}\right] \end{gathered}$$
As, $\text{AB}=\text{BC}=\text{AC}$ so the above equation becomes
$$\begin{gathered} \Rightarrow {\left(\text{AD}\right)}^2={\left(\text{AB}\right)}^2-\left[\frac{2{\left(\text{BC}\right)}^2}{9}\right] \\ \Rightarrow {\left(\text{AD}\right)}^2={\left(\text{AB}\right)}^2-\frac{2{\left(\text{AB}\right)}^2}{9} \\ \Rightarrow {\left(\text{AD}\right)}^2=\frac{9{\left(\text{AB}\right)}^2-2{\left(\text{AB}\right)}^2}{9} \\ \Rightarrow {\left(\text{AD}\right)}^2=\frac{7{\left(\text{AB}\right)}^2}{9} \\ \Rightarrow 9{\left(\text{AD}\right)}^2=7{\left(\text{AB}\right)}^2 \end{gathered}$$

Hence proved, $9A{D}^2=7A{B}^2$.