14. In the diagram, $A B=A C$ and $D$ is a point on $A C$ such that $B D=B C$.
Angle $B A C$ is $40^{\circ}$.
What is angle $A B D$ ?
A $15^{\circ}$
B $20^{\circ}$
C $25^{\circ}$
D $30^{\circ}$
E $35^{\circ}$
14. D As $A B=A C$, triangle $A B C$ is isosceles. So $\angle A B C=\angle A C B$ $=\frac{1}{2}\left(180^{\circ}-40^{\circ}\right)=70^{\circ}$ as $\angle B A C=40^{\circ}$ and the angle sum of a triangle is $180^{\circ}$. Triangle $B C D$ is also isosceles as $B D=B C$, so $\angle B D C=\angle B C D=70^{\circ}$. Considering triangle $A B D: \angle B D C=\angle D A B+\angle A B D$ as an exterior angle of a triangle is equal to the sum of the two interior opposite
angles. So $\angle A B D=\angle B D C-\angle D A B=70^{\circ}-40^{\circ}=30^{\circ}$.