Question

In the fig. 3.12 seg AD $\cong$ seg DC $\cong$ seg BD. Prove that $\angle$ABC is right angle.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{ABC}, \\ \mathrm{seg}\mathrm{AD}\cong \mathrm{seg}\mathrm{DC}\cong \mathrm{seg}\mathrm{BD},\mathrm{i}.\mathrm{e}.\mathrm{AD}=\mathrm{DC}=\mathrm{BD} \\ \mathrm{In}△\mathrm{ABD}, \\ \text{l}\mathrm{et}\angle \mathrm{DAB}=\angle \mathrm{ABD}=x(\because \mathrm{AD}=\mathrm{BD}) \\ \mathrm{In}△\mathrm{DBC}, \\ \text{l}\mathrm{et}\angle \mathrm{DBC}=\angle \mathrm{DCB}=\mathrm{y}(\because \mathrm{DC}=\mathrm{BD}) \\ \mathrm{In}△\mathrm{ABC}, \\ x+y+x+y\mathit{}=180° \\ \Rightarrow 2x+2y=180° \\ \Rightarrow x+y=90° \\ \mathrm{Thus},\angle \mathrm{B}=90°. \end{gathered}$$