Question

In fig. 3.5 and fig. 3.6 seg AB$\perp$ seg BC, seg DC $\perp$seg BC and seg AB $\cong$seg DC then show that seg BD $\cong$ seg AC

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{BDC}\mathrm{and}△\mathrm{BAC}, \\ \mathrm{CD}=\mathrm{AB}\left(\mathrm{given}\right) \\ \angle \mathrm{DCB}=\angle \mathrm{ABC}=90° \\ \mathrm{BC}=\mathrm{BC}\left(\mathrm{common}\right) \\ \therefore △\mathrm{BDC}\cong △\mathrm{BAC}(\text{b}y\mathrm{SAS}\mathrm{congruency}\mathrm{test}) \\ \mathrm{Thus},\mathrm{BD}=\mathrm{CA}(\text{by}\mathrm{cpct}) \\ \text{Or},\mathrm{seg}\mathrm{BD}\cong \mathrm{seg}\mathrm{AC} \end{gathered}$$