Question

In the figure below, $AB = AC = CD$. If $\angle ADB = 20^\circ$, what is the value of $\angle BAD$? (Triangle $ABD$ with point $C$ on $BD$ such that $AB=AC=CD$ and $\angle ADB = 20^\circ$)

Accepted Answer

D. In $	riangle ACD$, $AC = CD$, so $\angle CAD = \angle CDA = 20^\circ$. Exterior angle $\angle ACB = \angle CAD + \angle CDA = 20^\circ + 20^\circ = 40^\circ$. In $	riangle ABC$, $AB = AC$, so $\angle ABC = \angle ACB = 40^\circ$. Then $\angle BAC = 180^\circ - 40^\circ - 40^\circ = 100^\circ$. Hence $\angle BAD = \angle BAC + \angle CAD = 100^\circ + 20^\circ = 120^\circ$.

Geometry: Angle Calculation in Triangle

Explanation

Practise this under exam conditions