Question

The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.

Answer 1

Given parallelogram $ABCD$ in which diaginal $BD$ bisects $\angle B$ and $\angle D$

$BD$ bisect $\angle B,$

$\Rightarrow \angle ABD=\angle CBD=\frac{1}{2}\angle ABC$

$BD$ bisect $\angle D,$

$\Rightarrow \angle ADB=\angle CDB=\frac{1}{2}\angle ADC$

$\therefore \angle ABC=\angle ADC$ [ Opposite angles of parallelogram are equal ]

$\therefore \frac{1}{2}\angle ABC=\frac{1}{2}\angle ADC$

$\Rightarrow \angle ABD=\angle ADB$ and $\angle CBD=\angle CDB$

$\therefore AD=AB$ and $CD=BC$

As $ABCD$ is a parallelogram, opposite sides are equal

$\Rightarrow AB=CD$ and $AD=BC$

$\Rightarrow AB=BC=CD=AD$ i.e, $ABCD$ is a rhombus.

Hence, the answer is proved.