Question
ABCD is a rhombus. If $\angle\text{ACB}=40^\circ$, find $\angle\text{ADB}$.
✓
Answer
In a rhombus, the diagonals are perpendicular.$\therefore\angle\text{BPC}=90^\circ$
From $\triangle\text{BPC}$, the sum of angles is 180°.
$\therefore\angle\text{CBP}+\angle\text{BPC}+\angle\text{PBC}=180^\circ$
$\angle\text{CBP}=180^\circ-\angle\text{BPC}-\angle\text{PBC}$
$\angle\text{CBP}=180^\circ-40^\circ-90^\circ$
$\angle\text{CBP}=50^\circ$
$\angle\text{ADB}=\angle\text{CBP}=50^\circ$ (alternate angle)
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