Question

Prove that in a parallelogram, the bisects of any two consecutive angle intersect at right angle.

Answer 1

A parallelogram $ABCD$ such that the bisectors of adjacent angles $A$ and $B$ intersect at $O.$

$\Rightarrow$ $AD\parallel BC$

$\Rightarrow$ $\angle A+\angle B={180}^o$ [ Sum of consecutive interior angle ]

$\Rightarrow$ $\frac{1}{2}\angle A+\frac{1}{2}\angle B={90}^o$

$\Rightarrow$ $\angle OAB+\angle ABO={90}^o$ --- ( 1 )

In $△OAB,$

$\Rightarrow$ $\angle OAB+\angle ABO+\angle BOA={180}^o$

$\Rightarrow$ ${90}^o+\angle BOA={180}^o$ [ From ( 1 ) ]

$\therefore$ $\angle BOA={90}^o.$

Hence, we have prove that in a parallelogram, the bisects of any two consecutive angle intersect at right angle.