Question

In Figure, $ \mathrm{ABC}$ is a triangle in which $ \angle \mathrm{ABC}>90°$ and $ \mathrm{AD}\perp \mathrm{CB}$ produced. Prove that $ {\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+ {\mathrm{BC}}^{2}+ 2\mathrm{BC}.\mathrm{BD}.$

Answer 1

Prove that in the given figure ${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2+2\mathrm{BC}.\mathrm{BD}.$

Given: $\mathrm{AD}\perp \mathrm{CB}$

We know that Pythagoras theorem states that square of hypotenuse is equal to sum of the squares of other two sides in right triangle.

In $△\mathrm{ADB}$, ${\mathrm{AB}}^2={\mathrm{AD}}^2+{\mathrm{DB}}^2\mathbf{}(\mathrm{By}\mathrm{Pythagora}s\mathrm{theorem})$

In triangle $△\mathrm{ADC}$,

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AD}}^2+{\mathrm{DC}}^2(\mathrm{By}\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AD}}^2+{(\mathrm{DB}+\mathrm{DC})}^2(\because \mathrm{DC}=\mathrm{DB}+\mathrm{DC}) \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AD}}^2+{\mathrm{DB}}^2+{\mathrm{DC}}^2+2·\mathrm{DB}·\mathrm{DC}({(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}) \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{DC}}^2+2·\mathrm{DB}·\mathrm{DC}(\because {\mathrm{AB}}^2={\mathrm{AD}}^2+{\mathrm{DB}}^2) \end{gathered}$$

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