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