## Elementary Geometry for College Students (6th Edition)

by Theorem one diagonal of a kite bisects two angles of the kite which is $\angle A and \angle C$ therefore $\triangle ABC = \frac{1}{2} \angle A + \frac{1}{2} \angle C + \angle B= 180^{\circ}$ By substitution $(\angle B - 30) + (\angle B -50) + \angle B = 180$ $So \angle B = 110$
In the kite $ABCD$, the measures of $\angle B$ and $\angle D$ are equal. Also, the sum of the four interior angles in a kite is $360^{\circ}$ We can find $m\angle B$: $m\angle A + m\angle B + m\angle C + m\angle D = 360^{\circ}$ $(m\angle B - 50) + m\angle B + (m\angle B-30) + m\angle B = 360^{\circ}$ $4~m\angle B - 80 = 360^{\circ}$ $4~m\angle B = 440^{\circ}$ $m\angle B = \frac{440^{\circ}}{4}$ $m\angle B = 110^{\circ}$