Question

In the figure, D is a point on BC, such that $\angle \mathrm{ABD}=\angle \mathrm{CAD}$. AB=5 cm, AD=4 cm and AC=3 cm.
Find (i) BC
(ii) DC
(iii) A($△$ACD) : A($△$BCA)

Answer 1

Given : $\angle \mathrm{ABD}=\angle \mathrm{CAD}$
(i)
$$\begin{gathered} \text{In}△\mathrm{DAC}\text{and}△\mathrm{ABC}, \\ \angle \mathrm{DAC}=\angle \mathrm{ABC}\left(\mathrm{Given}\right) \\ \angle \mathrm{DCA}=\angle \mathrm{ACB}(\mathrm{Common}\mathrm{angle}) \\ \mathrm{Therefore},\mathrm{by}\mathrm{AA}\mathrm{similarity}\mathrm{test}, \\ △\mathrm{DAC}~△\mathrm{ABC} \\ \mathrm{Therefore},\frac{\mathrm{DA}}{\mathrm{AB}}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{DC}}{\mathrm{AC}}...\left(1\right) \\ \frac{\mathrm{DA}}{\mathrm{AB}}=\frac{\mathrm{AC}}{\mathrm{BC}} \\ \Rightarrow \frac{4}{5}=\frac{3}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{5\times 3}{4}=3.75\mathrm{cm} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{From}\left(1\right), \\ \frac{\mathrm{DA}}{\mathrm{AB}}=\frac{\mathrm{DC}}{\mathrm{AC}} \\ \Rightarrow \frac{4}{5}=\frac{\mathrm{DC}}{3} \\ \Rightarrow \mathrm{DC}=\frac{4\times 3}{5}=2.4\mathrm{cm} \end{gathered}$$

(iii)
$$\begin{gathered} △\mathrm{DAC}\&△\mathrm{ABC}\text{can be written}\mathrm{as}△\mathrm{ACD}\&△\mathrm{BCA},\mathrm{respectively}. \\ \mathrm{Therefore},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}△\mathrm{ACD}~△\mathrm{BCA}. \\ \mathrm{Hence},\frac{\mathrm{A}(△\mathrm{ACD})}{\mathrm{A}(△\mathrm{BCA})}=\frac{{\mathrm{AC}}^2}{{\mathrm{BC}}^2}=\frac{{\mathrm{CD}}^2}{{\mathrm{CA}}^2}=\frac{{\mathrm{DA}}^2}{{\mathrm{AB}}^2} \\ \Rightarrow \frac{\mathrm{A}(△\mathrm{ACD})}{\mathrm{A}(△\mathrm{BCA})}=\frac{{\mathrm{DA}}^2}{{\mathrm{AB}}^2} \\ \Rightarrow \frac{\mathrm{A}(△\mathrm{ACD})}{\mathrm{A}(△\mathrm{BCA})}=\frac{4^2}{5^2} \\ \Rightarrow \frac{\mathrm{A}(△\mathrm{ACD})}{\mathrm{A}(△\mathrm{BCA})}=\frac{16}{25} \\ \mathrm{A}(△\mathrm{ACD}):\mathrm{A}(△\mathrm{BCA})=16:25 \end{gathered}$$