ABCD is a rhombus.


ABCD is a rhombus. If ∠ACB = 40°, find ∠ADB.


In a rhombus, the diagonals are perpendicular.

$\therefore \angle \mathrm{BPC}=90^{\circ}$

From $\Delta$ BPC, the sum of angles is $180^{\circ}$.

$\therefore \angle \mathrm{CB} P+\angle \mathrm{BP} C+\angle P B C=180^{\circ}$

$\angle \mathrm{CB} P=180^{\circ}-\angle \mathrm{BP} C-\angle P B C$

$\angle \mathrm{CBP}=180^{\circ}-40^{\circ}-90^{\circ}=50^{\circ}$

$\angle A D B=\angle \mathrm{CBP}=50^{\circ}$ (alternate angle)

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