Solution
D

We suppose that we have chosen units so that the length of $G H$ is 1.

Because $F G H$ is a $60^{\circ}, 30^{\circ}, 90^{\circ}$ triangle, it follows that $F G$ has length $\frac{1}{2}$.
Because $E G H$ is a right-angled isosceles triangle it also follows that $E H$ has length $\frac{1}{\sqrt{2}}$.
In the triangles $I F G$ and $I E H$ we have
$$
\angle G F I=\angle H E I=90^{\circ}
$$
and
$$
\angle G I F=\angle H I E \text { (vertically opposite angles). }
$$

Because the sum of the angles in both these triangles is $180^{\circ}$, it follows that
$$
\angle F G I=\angle E H I .
$$

Therefore the triangles $I F G$ and $I E H$ are similar.
The ratio of the areas of similar triangles equals the ratio of the squares of the lengths of corresponding sides. Therefore
$$
\text { area of } I F G \text { : area of } \begin{aligned}
I E H & =F G^{2}: E H^{2} \\
& =\left(\frac{1}{2}\right)^{2}:\left(\frac{1}{\sqrt{2}}\right)^{2} \\
& =\frac{1}{4}: \frac{1}{2} \\
& =1: 2 .
\end{aligned}
$$
19. The diagram shows two overlapping triangles: triangle $F G H$ with interior angles $60^{\circ}, 30^{\circ}$ and $90^{\circ}$ and triangle $E G H$ which is a right-angled isosceles triangle.
What is the ratio of the area of triangle $I F G$ to the area of triangle $I E H$ ?
A $1: 1$
B $1: \sqrt{2}$
C $1: \sqrt{3}$
D $1: 2$
E $1: 3$