Question

In the fig. 3.17 seg AB $\perp$ seg BC, seg DC $\perp$ seg BC and seg AB $\cong$ segDC show that, seg AC and seg BD bisect each other at O.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{AB}=\mathrm{DC}, \\ \mathrm{DC}\perp \mathrm{BC}\text{and}\mathrm{AB}\perp \mathrm{BC}. \\ \mathrm{In}△\mathrm{ABC}\mathrm{and}△\mathrm{DBC}, \\ \mathrm{AB}=\mathrm{DC}\left(\mathrm{given}\right) \\ \angle \mathrm{B}=\angle \mathrm{C}=90° \\ \mathrm{BC}=\mathrm{CB}\left(\mathrm{common}\right) \\ \mathrm{Thus},△\mathrm{ABC}\cong △\mathrm{DCB}(\mathrm{SAS}\mathrm{test}) \\ △\mathrm{ABC}-△\mathrm{BOC}\cong △\mathrm{DCB}-△\mathrm{BOC} \\ \therefore △\mathrm{AOB}\cong △\mathrm{COD} \\ \mathrm{seg}\mathrm{AO}\cong \mathrm{seg}\mathrm{CO}\mathrm{and}\mathrm{seg}\mathrm{BO}\cong \mathrm{seg}\mathrm{DO}(\mathrm{by}\mathrm{c}.\mathrm{s}.\mathrm{c}.\mathrm{t}) \\ \mathrm{Hence},\mathrm{AC}\mathrm{and}\mathrm{BD}\mathrm{bisect}\mathrm{each}\mathrm{other}. \end{gathered}$$