Question

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

• A. BC || ED
• B. $\angle CBD={40}^{\circ }$
• C. $\angle CBD={90}^{\circ }$
• D. $\angle BDE={60}^{\circ }$

Answer 1

Answer: (A), (B)

The correct options are
A BC || ED
B $\angle CBD={40}^{\circ }$

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