Question

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

Answer 1

$ABCD$ is a parallelogram, the bisector of $\angle ADC$ and $\angle BCD$ meet at point $E$ and the bisectors of $\angle BCD$ and $\angle ABC$ meet at $F$. We have to prove that the $\angle CED={90}^o$ and $\angle CFG={90}^o$. This way we will be able to prove that $DE$ and $CE$ intersect at right angles and $BG$ and $ED$ are parallel. This way we will be able to prove that $DE$ and $CE$ intersect at right angles and $BG$ and $ED$ are parallel.

$\angle ADC+\angle BCD={180}^o$ [sum of adjacent angles of a parallelogram]

$\Rightarrow \frac{\angle ADC}{2}+\frac{\angle BCD}{2}={90}^o$

$\therefore \angle EDC+\angle ECD={90}^o$

In $△ECD$ sum of angle $={180}^o$

$\Rightarrow \angle EDC+\angle ECD+\angle CED={180}^o$

$\therefore \angle CED={90}^o$

Hence, the first condition is proved that in a parallelogram the bisectors of angles intersect at ${90}^o$

In $△ECD$, sum of angles $={180}^o$

$\Rightarrow \angle EDC+\angle ECD+\angle CED={180}^o$

$\angle CED={90}^o$

Hence, the first condition is proved that in a parallelogram, the bisectors of angle intersect at ${90}^o$

Similarly taking $ABCF,\angle BFC={90}^o$

$\angle BFC+\angle CFG={180}^o$

$\angle CFG={90}^o$

$\therefore DE||BG$

Hence proved