Question

In fig. 3.16 seg AB $\cong$ seg DC, $\angle$ABC $\cong$ $\angle$DCB show that seg AC $\cong$ seg BD and $\angle$A$\cong$ $\angle$D.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{ABC}\mathrm{and}△\mathrm{DCB}, \\ \mathrm{AB}=\mathrm{DC}\left(\mathrm{given}\right) \\ \angle \mathrm{ABC}=\angle \mathrm{DCB}\left(\mathrm{given}\right) \\ \mathrm{BC}=\mathrm{CB}\left(\mathrm{common}\right) \\ \mathrm{Thus},△\mathrm{ABC}\cong △\mathrm{DCB}(\mathrm{SAS}\mathrm{test}) \\ \therefore \mathrm{AC}=\mathrm{BD}(\mathrm{by}\mathrm{c}.\mathrm{s}.\mathrm{c}.\mathrm{t}) \\ \text{A}\mathrm{nd},\angle \mathrm{A}=\angle \mathrm{D}(\mathrm{by}\mathrm{c}.\mathrm{a}.\mathrm{c}.\mathrm{t}) \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$