2009-II-49 – Solving Master

Ans: C
Let

∠ C B D = x

. Since

A

B

C

and

D

are four points on the circumference, then

$\begin{array}{rcl} ∠ A C D & = & ∠ A B D \\ = & 40^{\circ} \end{array}$

Since $A B / / D C$ , then

$\begin{array}{rcl} ∠ B A C & = & ∠ A C D \\ = & 40^{\circ} \end{array}$

In $Δ A B C$ , since $A B = A C$ , then

$\begin{array}{rcl} ∠ A C B & = & ∠ A B C \\ = & 40^{\circ} + x \end{array}$

Hence, we have

$\begin{array}{rcl} 40^{\circ} + (40^{\circ} + x) \times 2 & = & 180^{\circ} \\ 120^{\circ} + 2 x & = & 180^{\circ} \\ 2 x & = & 60^{\circ} \\ x & = & 30^{\circ} \end{array}$

Therefore, $∠ C B D = 30^{\circ}$ .

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