Question

D is a point on side BC of $△$ABC, such that $\angle \mathrm{ADC}=\angle \mathrm{BAC}$. Show that AC² = BC$\times$DC.

Answer 1

Given : $\angle \mathrm{ADC}=\angle \mathrm{BAC}$

To prove : AC² = BC$\times$DC

Proof :

$$\begin{gathered} \text{In}△\mathrm{ADC}\text{and}△\mathrm{BAC}, \\ \angle \mathrm{ADC}=\angle \mathrm{BAC}(\mathrm{Given}) \\ \angle \mathrm{DCA}=\angle \mathrm{ACB}(\mathrm{Common}\mathrm{angle}) \\ \mathrm{Therefore},\mathrm{by}\text{the}\mathrm{AA}\mathrm{similarity}\mathrm{test}, \\ △\mathrm{ADC}~△\mathrm{BAC} \\ \mathrm{By}\text{the}\mathrm{property}\mathrm{of}\mathrm{similarity}, \\ \frac{\mathrm{AD}}{\mathrm{BA}}=\frac{\mathrm{DC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{BC}} \\ \Rightarrow \frac{\mathrm{DC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{BC}} \\ \Rightarrow {\mathrm{AC}}^2=\mathrm{DC}\times \mathrm{BC} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$