Question

In Fig. 7.243, if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, then BD = ________.

Answer 1

In ∆ABC and ∆ADC,

∠A = ∠A (Common)

∠ACB = ∠CDA (Given)

∴ ∆ABC $~$ ∆ACD (AA similarity criterion)

$\Rightarrow \frac{\mathrm{BC}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{CA}}=\frac{\mathrm{CA}}{\mathrm{AD}}$ (If two triangles are similar, then their corresponding sides are proportional)

$$\begin{gathered} \Rightarrow \frac{\mathrm{AB}}{\mathrm{CA}}=\frac{\mathrm{CA}}{\mathrm{AD}} \\ \Rightarrow \frac{3+\mathrm{BD}}{8}=\frac{8}{3} \\ \Rightarrow 9+3\mathrm{BD}=64 \\ \Rightarrow 3\mathrm{BD}=64-9=55 \end{gathered}$$
$\Rightarrow \mathrm{BD}=\frac{55}{3}\mathrm{cm}$

In Fig. 7.243, if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, then BD = $\underline{\frac{55}{3}\mathrm{cm}}$.