Question

In case of a parallelogram prove that :
(i) the bisector of any two adjacent angles intersect at ${90}^o$.
(ii) the bisector of opposite angle are parallel to each other.

Answer 1

i) Given $\square ABCD$ is a parallelogram

To prove : $\angle AOB={90}^o$

Proof:

$\angle DAB+\angle ABC={180}^o\to$ [ corresponding Interior angles]

$\frac{1}{2}\angle DAB+\frac{1}{2}=\angle ABC={90}^o$ [Multiplying $\frac{1}{2}$ on both sides]

$\angle DAB+\angle OBA={90}^o\to \left(1\right)$

In $△AOB$

$\angle AOB+\angle OAB+\angle OBA={180}^o\to$ [Angle $\delta m$ property]

$\angle AOB+{90}^o={180}^o\to$ [from $\left(1\right)$]

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

Hence proved

ii)

Given $AE$ is the bisector of $\angle DAB$

$CF$ is the bisector of $\angle DCB$

To prove: $AE\parallel CF$

Proof:

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

$\frac{1}{2}\angle A=\frac{1}{2}\angle C\to$ [ Multiply by $\frac{1}{2}$]

$\angle EAB=\angle DCF\to$ [$AE$ & $CF$ are bisector](1)

$\angle DCF=\angle CFB\to$ [ Alternate interior angles are equal] (2)

$\therefore \angle EAB=\angle CFB\to$ [ from $\left(1\right)$ and $\left(2\right)$]

$\therefore AB\parallel CF\to$ [ Corresponding angles are equal]

Hence proved