In a kite $ABCD$ with $AB = AD$ and $CB = CD$, $\angle B = 100^{\circ}$. Then $\angle D$ is:

In a kite $ABCD$ with $AB = AD$ and $CB = CD$, $\angle B = 100^{\circ}$. Then $\angle D$ is:
  1. $80^{\circ}$
  2. $90^{\circ}$
  3. $100^{\circ}$
  4. $160^{\circ}$

Solution

In a kite the angles between the pairs of unequal sides are equal. So $\angle B = \angle D = 100^{\circ}$.

Asked in: IMO

Practice more UNDERSTANDING QUADRILATERALS questions on Aicharya