2. In the square $A B C D$, the bisector of $\angle C A D$ meets $C D$ at $P$ and the bisector of $\angle A B D$ meets $A C$ at $Q$.
What is the ratio of the area of triangle $A C P$ to the area of triangle $B Q A$ ?
We know that, since $A C$ is the diagonal of a square, $\angle P C A=45^{\circ}=\angle Q A B$. Also $\angle D A C=$ $45^{\circ}=\angle D B A$. Since $A P$ and $B Q$ are angle bisectors, $\angle P A C=22.5^{\circ}=\angle Q B A$.
Hence the triangles $C P A$ and $A Q B$ have equal angles, so they are similar. Now $C A=\sqrt{2}$ and $A B=1$, so the sides are in the ratio $\sqrt{2}: 1$, and the areas are in the ratio $2: 1$.