Question

ln $△\mathrm{ABC}$, $\mathrm{AD}\perp \mathrm{BC}$ and point $D$ lies on $BC$ such that $2\mathrm{BD}=3\mathrm{CD}$. Prove that $5{\mathrm{AB}}^2=5{\mathrm{AC}}^2+{\mathrm{BC}}^2$.

Answer 1

Prove the given equation.

In the question, it is given that in $△\mathrm{ABC}$, $\mathrm{AD}\perp \mathrm{BC}$ and point $D$ lies on $BC$ such that $2\mathrm{BD}=3\mathrm{CD}$.

Assume that, $\mathrm{BD}=3\mathrm{x}$. So, $\mathrm{CD}=2\mathrm{x}$.

$$\begin{gathered} \mathrm{BC}=\mathrm{BD}+\mathrm{CD} \\ \Rightarrow \mathrm{BC}=5\mathrm{x} \end{gathered}$$

In $△\mathrm{ABD}$ by Pythagoras theorem.

$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{BD}}^2+{\mathrm{AD}}^2 \\ \Rightarrow {\mathrm{AD}}^2={\mathrm{AB}}^2-9{\mathrm{x}}^2...\left(i\right) \end{gathered}$$

In $△\mathrm{ADC}$ by Pythagoras theorem.

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{CD}}^2+{\mathrm{AD}}^2 \\ \Rightarrow {\mathrm{AD}}^2={\mathrm{AC}}^2-4{\mathrm{x}}^2...\left(ii\right) \end{gathered}$$

From equation $\left(i\right)$ and $\left(ii\right)$

$$\begin{gathered} {\mathrm{AC}}^2-4{\mathrm{x}}^2={\mathrm{AB}}^2-9{\mathrm{x}}^2 \\ \Rightarrow {\mathrm{AC}}^2+5{\mathrm{x}}^2={\mathrm{AB}}^2 \end{gathered}$$

Multiply both sides by $5$.

$$\begin{gathered} \Rightarrow 5{\mathrm{AC}}^2+25{\mathrm{x}}^2=5{\mathrm{AB}}^2 \\ \Rightarrow 5{\mathrm{AC}}^2+{\mathrm{BC}}^2=5{\mathrm{AB}}^2\left[\because \mathrm{BC}=5\mathrm{x}\right] \end{gathered}$$

Hence proved $5{\mathrm{AB}}^2=5{\mathrm{AC}}^2+{\mathrm{BC}}^2$.