7. Let $D$ be a point inside triangle $\triangle A B C$ such that $A B=D C, \angle D C A=24^{\circ}, \angle D A C=31^{\circ}$ and $\angle A B C=55^{\circ}$. Find $\angle D A B$.
(1 mark)
設 $D$ 為 $\triangle A B C$ 內的一點, 使得 $A B=D C, \angle D C A=24^{\circ}, \angle D A C=31^{\circ}$, $\angle A B C=55^{\circ}$ 。求 $\angle D A B$ 。
7. $63^{\circ}$
7. Draw $E$ such that $A B C E$ is a parallelogram.
Since $\angle A E C=\angle A B C=55^{\circ}$ and $\angle A D C=$ $180^{\circ}-31^{\circ}-24^{\circ}=125^{\circ}$, we have $\angle A E C+$ $\angle A D C=180^{\circ}$ and thus $A D C E$ is a cyclic quadrilateral. Now $E C=A B=D C$, so $\angle C D E=$ $\angle C E D=\angle C A D=31^{\circ}$. Considering $\triangle C D E$, we have $\angle A C E=180^{\circ}-31^{\circ}-31^{\circ}-24^{\circ}=94^{\circ}$. It follows that $\angle D A B=\angle B A C-31^{\circ}=\angle A C E-$ $31^{\circ}=94^{\circ}-31^{\circ}=63^{\circ}$.
(Alternatively, instead of drawing the point $E$, one can reflect $B$ across $A C$ and proceed in essentially the same way.)