4. The diagram shows the two squares $B C D E$ and $F G H I$ inside the triangle $A B J$, where $E$ is the midpoint of $A B$ and $C$ is the midpoint of $F G$.
What is the ratio of the area of the square $B C D E$ to the area of the triangle $A B J$ ?
Solution
Let $K$ be the foot of the perpendicular from $J$ to $A B$ produced, as shown alongside.
Since $E$ is the midpoint of $A B$, we have $A E=E B=E D$ and so $\angle D A E=\angle J A K=45^{\circ}$.
Let $F G=2 y$ and $B K=x$. Then $F C=C G=y, D F=F I=2 y$ and $D C=B C=3 y$.
Triangles $G C B$ and $B K J$ are similar, since their angles are equal.
Hence $\frac{J K}{K B}=\frac{C B}{C G}=3$ and so $J K=3 x$.
But $J K=A K=6 y+x$. It follows that $3 x=6 y+x$ and so $x=3 y$.
The area of $B C D E$ is $9 y^{2}$ and that of $A B J$ is $27 y^{2}$, so the required ratio is $1: 3$.