Question

In the figure below, $△$ABC $\cong$ $△$BED, $\angle$DEB = ${115}^{\circ }$ and $\angle$CAB = ${25}^{\circ }$. It can be concluded that BC||ED.

• A. True
• B. False

Answer 1

The correct option is A

True

Since $\Delta ABC\cong \Delta BED,\angle DBE=\angle CAB={25}^{\circ }\text{(CPCT)}\text{And,}\angle DEB=\angle CBA\text{(CPCT)}\text{Now,}\angle BDE+\angle DBE+\angle BED={180}^{\circ }\Rightarrow \angle BDE+{25}^{\circ }+{115}^{\circ }={180}^{\circ }\Rightarrow \angle BDE={40}^{\circ }\text{Also,}\angle CBA+\angle CBD+\angle DBE={180}^{\circ }\text{(Linear set of angles)}\Rightarrow \angle CBA+{25}^{\circ }+{115}^{\circ }={180}^{\circ }\Rightarrow \angle CBA={40}^{\circ }\Rightarrow \angle BDE=\angle CBD\text{and they are alternate angles, so}BC||ED.$