Question : $\triangle ABC$ is an equilateral triangle. The side $BC$ is produced to point $D$. If $A$ joines $D$ and $B C=CD$, then the degree measure of angle $CAD$ is equal to:
Option 1: $30^{\circ}$
Option 2: $15^{\circ}$
Option 3: $45^{\circ}$
Option 4: $18^{\circ}$
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Correct Answer: $30^{\circ}$
Solution : Given: BC = CD And ABC is an equilateral triangle. $\therefore$ AB = BC = CA $\angle ACB = \angle ABC = \angle BAC = 60°$ To find $\angle CAD$ $\angle ACB + \angle ACD = 180°$ ⇒ $ \angle ACD = 180°-60°$ ⇒ $\angle ACD = 120°$ $\triangle ACD$ is an isosceles triangle $\therefore$ $\angle CAD = \angle ADC$ $\angle CAD + \angle ACD + \angle ADC = 180°$ ⇒ $\angle CAD + 120° + \angle ACD = 180°$ ⇒ $2\angle CAD = 60°$ ⇒ $\angle CAD = 30°$ Hence, the correct answer is $30^{\circ}$.
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