In a rhombus ABCD,angle B = 80^{circ},then an | vedclass

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(D) In a rhombus $ABCD$,all sides are equal,so $AB = AD$. Since $AB = AD$,$	riangle ABD$ is an isosceles triangle. In a rhombus,adjacent angles are s

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$40$

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(D) In a rhombus $ABCD$,all sides are equal,so $AB = AD$. Since $AB = AD$,$	riangle ABD$ is an isosceles triangle. In a rhombus,adjacent angles are supplementary,so $\angle A + \angle B = 180^{\circ}$. Given $\angle B = 80^{\circ}$,then $\angle A = 180^{\circ} - 80^{\circ} = 100^{\circ}$. In $	riangle ABD$,the sum of angles is $180^{\circ}$,so $\angle A + \angle ADB + \angle ABD = 180^{\circ}$. Since $AB = AD$,the base angles are equal: $\angle ADB = \angle ABD$. Therefore,$100^{\circ} + 2(\angle ADB) = 180^{\circ}$. $2(\angle ADB) = 80^{\circ}$. $\angle ADB = 40^{\circ}$.

In a rhombus $ABCD$,$\angle B = 80^{\circ}$,then $\angle ADB = \ldots$ (in $^{\circ}$)

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