Question 50
In the figure below, AB = AC = CD. If ADB = 20°, what is the value of BAD?
Solution
AB = AC = CD, => $$\angle CAD = \angle CDA = 20^{\circ}$$
and $$\angle ABC = \angle ACB$$
In $$\triangle$$ ACD
=> $$\angle ACD + \angle CAD + \angle CDA = 180^{\circ}$$
=> $$\angle ACD = 180^{\circ} - 20^{\circ} - 20^{\circ} = 140^{\circ}$$
=> $$\angle ACB = 180^{\circ} - 140^{\circ} = 40^{\circ} = \angle ABC$$
Similarly, In $$\triangle$$ ABC
=> $$\angle BAC = 180^{\circ} - 40^{\circ} - 40^{\circ} = 100^{\circ}$$
$$\therefore \angle BAD = 100^{\circ} + 20^{\circ} = 120^{\circ}$$
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