H5. In the trapezium $A B C D$, the lines $A B$ and $D C$ are parallel, $B C=A D, D C=2 A D$ and $A B=3 A D$.
The angle bisectors of $\angle D A B$ and $\angle C B A$ intersect at the point $E$.
What fraction of the area of the trapezium $A B C D$ is the area of the triangle $A B E$ ?
Solution
Let $B C=A D=k$, so that $D C=2 k$ and $A B=3 k$, and let the point $X$ lie on $A B$ so that $X B C D$ is a parallelogram, as shown in the diagram on the left below. It follows that $D X=k$ and $X B=2 k$ (opposite sides of a parallelogram), so that $A X=k$.
Hence triangle $A X D$ has three equal sides-it is therefore an equilateral triangle. In particular, this means that angle $D A X$ is equal to $60^{\circ}$.
As a consequence, the trapezium $A B C D$ is actually made up from five equilateral triangles, as shown in the diagram on the right above.
Now the triangle $A B E$ comprises one equilateral triangle and two half-rhombuses. The area of the two half-rhombuses is equal to the area of two equilateral triangles.
Therefore the area of the triangle $A B E$ is $\frac{3}{5}$ of the area of the trapezium $A B C D$.