Question

∆ABC is a right triangle right-angled at A and AD ⊥ BC. Then, $\frac{\mathrm{BD}}{\mathrm{DC}}=$
(a) ${\left(\frac{\mathrm{AB}}{\mathrm{AC}}\right)}^2$
(b) $\frac{\mathrm{AB}}{\mathrm{AC}}$
(c) ${\left(\frac{\mathrm{AB}}{\mathrm{AD}}\right)}^2$
(d) $\frac{\mathrm{AB}}{\mathrm{AD}}$

Answer 1

Given: In ΔABC, and.

To find: BD: DC

$$\begin{gathered} \angle \mathrm{CAD}+\angle \mathrm{BAD}=90°.....\left(1\right) \\ \angle \mathrm{BAD}+\angle \mathrm{ABD}=90°.....\left(2\right)\left(\angle \mathrm{ADB}=90°\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right), \\ \angle \mathrm{CAD}=\angle \mathrm{ABD} \end{gathered}$$

In ΔADB and ΔADC,

$$\begin{gathered} \angle \mathrm{ADB}=\angle \mathrm{ADC}\left(90°\mathrm{each}\right) \\ \angle \mathrm{ABD}=\angle \mathrm{CAD}\left(\mathrm{Proved}\right) \\ \therefore ∆\mathrm{ADB}~∆\mathrm{ADC}\left(\mathrm{AA}\mathrm{Similarity}\right) \\ \Rightarrow \frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{AD}}{\mathrm{BD}}\left(\mathrm{Corresponding}\mathrm{sides}\mathrm{are}\mathrm{proportional}\right) \end{gathered}$$

Disclaimer: The question is not correct. The given ratio cannot be evaluated using the given conditions in the question.