Question

In triangle $\mathrm{ABC}$, $\mathrm{AD}$ is perpendicular to $\mathrm{BC}$ and ${\mathrm{AD}}^2=\mathrm{BD}.\mathrm{CD}$.

Prove that $△\mathrm{ABC}$ is a right triangle.

Answer 1

Representing the situation Diagrammatically

In the question, it is given that in triangle $\mathrm{ABC}$, $\mathrm{AD}$ is perpendicular to $\mathrm{BC}$ and ${\mathrm{AD}}^2=\mathrm{BD}.\mathrm{CD}$.

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

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

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

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

Add equation $\left(i\right)$ and equation $\left(ii\right)$.

$$\begin{gathered} 2{\mathrm{AD}}^2={\mathrm{AB}}^2-{\mathrm{BD}}^2+{\mathrm{AC}}^2-{\mathrm{CD}}^2\left[\because {\mathrm{AD}}^2=\mathrm{BD}.\mathrm{CD}\left(\mathrm{given}\right)\right] \\ \Rightarrow 2\mathrm{BD}.\mathrm{CD}+{\mathrm{CD}}^2+{\mathrm{BD}}^2={\mathrm{AB}}^2+{\mathrm{AC}}^2\left[\because {\left(\mathrm{a}+\mathrm{b}\right)}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\right] \\ \Rightarrow {\left(\mathrm{BD}+\mathrm{CD}\right)}^2={\mathrm{AB}}^2+{\mathrm{AC}}^2\left[\because \mathrm{BC}=\mathrm{BD}+\mathrm{CD}\right] \\ \Rightarrow {\mathrm{BC}}^2={\mathrm{AB}}^2+{\mathrm{AC}}^2 \end{gathered}$$

By converse of Pythagoras theorem $△\mathrm{ABC}$ is right angled at $A$.

Hence Proved $△\mathrm{ABC}$ is a right triangle.