ABCD is a rhombus. Show that diagonal AC bisects $\angle A$as well as $\angle C$and diagonal BD bisects $\angle B$as well as $\angle D\dot{}$

Solution: We can construct a diagram with the given details.
Please refer to video for the diagram.
Now, in $\Delta AEB\text{and}\Delta AEC$,
$AB=AC$(sides of rhombus)
$\angle AEB=\angle AEC={90}^{\circ }$
$AE=AE$(Common sides)
So by, RHS criterion,
$\Delta AEB\stackrel{\sim }{=}\Delta AEC$
It means,$\angle BAE=\angle DAE$ or $\angle BAC=\angle DAC$
It implies, diagonal $AC$ bisects $\angle A$.
Similarly, diagonal $AC$ bisects $\angle C$.
Similarly, we can show that diagonal $BD$ bisects $\angle B$ as well as $\angle D$.