Question

Prove that the bisectors of two opposite angles of a parallelogram are parallel.

Answer 1

In the figure shown above, $ABCD$ is a parallelogram.

$\therefore \angle A=\angle C$ [Opposite angles of parallelogram are equal]

$AX$ and $CY$ are bisectors of $\angle A$ and $\angle C$ respectively.

Hence, $\frac{1}{2}\angle A=\frac{1}{2}\angle C$

$\Rightarrow$ $\angle 1=\angle 2$ ---- ( 1 )

Now, $AB\parallel DC$ and the transversal $CY$ intersects them.

$\therefore \angle 2=\angle 3$ ---- ( 2 ) [$\because$ Alternate interior angles are equal]

From ( 1 ) and ( 2 ), we get:

$\angle 1=\angle 3$

Thus, transversal $AB$ intersects $AX$ and $YC$ at $A$ and $Y$ such that $\angle 1=\angle 3$, that is, corresponding angles are equal.

$\therefore AX\parallel CY$

Hence, it is proved that the bisectors of two opposite angles of a parallelogram are parallel.