Question

In the figure given below, prove that $p\parallel m$.

Answer 1

In $\Delta BCD$,

$\angle CBD+\angle BCD+\angle BDC={180}^{\circ }$

[Since the sum of the angles of a triangle is ${180}^{\circ }$]

$\Rightarrow \angle CBD+{45}^{\circ }+{35}^{\circ }={180}^{\circ }$

$\Rightarrow \angle CBD+{80}^{\circ }={180}^{\circ }$

$\Rightarrow \angle CBD={180}^{\circ }-{80}^{\circ }={100}^{\circ }$ ....(i)

$\Rightarrow \angle EBD+\angle CBD={180}^{\circ }$ (Linear pair of angles)

$\Rightarrow \angle EBD+{100}^{\circ }={180}^{\circ }$ ......(From i)

$\Rightarrow \angle EBD={80}^{\circ }$

$\Rightarrow \angle EBD=\angle FAB$ (corresponding angles)

These angles form an equal pair of corresponding angles for lines p and m with n as transversal. Therefore it can be concluded that p is parallel to m.