Question

In the figure, p is a transversal to lines m and n, $\angle 2={120}^{0}and\angle 5={60}^0$. Prove that m||n

Answer 1

Given p is a transversal to the lines m and n
Forming $\angle 1,\angle 2,\angle 3,\angle 4,\angle 5,\angle 6,\angle 7and\angle 8$
$\angle 2={120}^0,and\angle 5={60}^0$
To prove: m||n
Proof: $\angle 2+\angle 3={180}^0$
$\Rightarrow {120}^0+\angle 3={180}^0$
$\Rightarrow \angle 3={180}^0-{120}^0={60}^0$
But $\angle 5={60}^0$
$\therefore \angle 3=\angle 5={60}^0$
$\therefore \angle 3=\angle 5$
But there are alternate angles
$\therefore m||n$