19. The square $A B C D$ has sides of length 105 . The point $M$ is the midpoint of side $B C$. The point $N$ is the midpoint of $B M$. The lines $B D$ and $A M$ meet at the point $P$. The lines $B D$ and $A N$ meet at the point $Q$.
What is the area of triangle $A P Q$ ?
SolUTION
735

Let $V$ be the centre of the square $A B C D$. Let $W$ be the intersection between $B D$ and the line through $N$ parallel to $A B$.
Triangles $A P B$ and $M P V$ are similar, with $B P: P V=A B$ : $M V=\frac{1}{2}$. Therefore, $B P=\frac{2}{3} \times B V=\frac{2}{3} \times \frac{1}{2} \times 105 \sqrt{2}=35 \sqrt{2}$. Similarly, triangles $A Q B$ and $N Q W$ are similar, with $B Q: Q W=$ $A B: N W=\frac{1}{4}$. Therefore, $B Q=\frac{4}{5} \times B W=\frac{4}{5} \times \frac{1}{4} \times 105 \sqrt{2}=$ $21 \sqrt{2}$
The area of $A P Q$ is $\frac{1}{2} \times Q P \times V A=\frac{1}{2} \times(35 \sqrt{2}-21 \sqrt{2}) \times \frac{1}{2} \times$ $105 \sqrt{2}=\frac{1}{2} \times 14 \sqrt{2} \times \frac{1}{2} \times 105 \sqrt{2}=735$.