Question

$△\mathrm{ABC}$ is inscribed in a circle with centre O, seg AX is a diameter of the circle with radius r, seg AD$\perp$ seg BC
Prove that:-
$$\begin{gathered} \left(\mathrm{i}\right)△\mathrm{ABX}~△\mathrm{ADC} \\ \left(\mathrm{ii}\right)\mathrm{A}(△\mathrm{ABC})=\frac{\mathit{a}\mathit{b}\mathit{c}}{\mathit{4}\mathit{r}} \\ (\mathrm{a}\mathrm{is}\mathrm{side}\mathrm{opposite}\mathrm{to}\angle \mathrm{A}) \end{gathered}$$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{In}△\mathrm{ABX}\mathrm{and}∆\mathrm{ADC},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{ABX}=\angle \mathrm{ADC}=90°(\because \mathrm{AX}\mathrm{is}\mathrm{a}\mathrm{diameter}\mathrm{so},\angle \mathrm{ABX}=90°) \\ \angle \mathrm{AXB}=\angle \mathrm{ACB}(\mathrm{Angles}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{segment}) \\ \therefore △\mathrm{ABX}~∆\mathrm{ADC}(\mathrm{By}\mathrm{AA}\mathrm{similarity}) \\ \mathrm{i}.\mathrm{e}.,\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AX}}{\mathrm{AC}} \\ \Rightarrow \mathrm{AD}=\frac{\mathrm{AB}\times \mathrm{AC}}{\mathrm{AX}}...\left(1\right) \\ \left(\mathrm{ii}\right) \\ \mathrm{AB}=c \\ \mathrm{AC}=b \\ \mathrm{BC}=a \\ F\mathrm{rom}\left(1\right) \\ \mathrm{AD}=\frac{c\mathit{\times }b}{2r} \\ \therefore \mathrm{A}(△\mathrm{ABC})=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ =\frac{1}{2}\times \mathrm{BC}\times \mathrm{AD} \\ =\frac{1}{2}\times a\times \frac{c\mathit{\times }b}{2r} \\ =\frac{abc}{4r} \end{gathered}$$