A trapezium inside a triangle. This question was asked in a scholarship test for class 10th.
In the given figure, ABCD is a trapezium, AD is parallel to BC and diagonal BD is parallel to XY. 3AD=BC. Find the ratio of AY to AX.

My attempt (By using Similarity of Triangles):-
YD/DC=YA/AX
YD/DC=BX/BC
YA/AX=BX/BC
Let AD=y, BC=3y.
On producing AB and CY, let them intersect at point T.
TA/AB=TD/DC=AD/BC=1/3
So, AB=2TA and CD=2TD
 A: It's unnecessary to extend $AB$ and $CY$ until they intersect at a point $T$. Instead, here's your diagram with an extra line of $BD$, and a few lengths, added to it:

In particular, setting $|AX|=e$ and $\frac{|AY|}{|AX|}=r$, then $|AY|=re$. With $AD \parallel BC$, we get $\triangle YAD \sim \triangle YXC$. Thus, with $|DC|=g$, we have $|YD|=rg$.
Setting $|AD|=f$ (instead of your $y$, to help avoid any possible confusion with $Y$), then $|BC| = 3|AD| = 3f$. Next, as you've already noted, with $BD \parallel XY$, then $\triangle CBD \sim \triangle CXY$, so $|BX|=r|BC|=3rf$.
Using $\triangle YAD \sim \triangle YXC$ again, then
$$\begin{equation}\begin{aligned}
\frac{|YA|}{|AD|} & = \frac{|YX|}{|XC|} \\
\frac{re}{f} & = \frac{(r+1)e}{3(r+1)f} \\
\frac{re}{f} & = \frac{e}{3f} \\
r & = \frac{1}{3}
\end{aligned}\end{equation}$$
Thus, $|AY|:|AX|=1:3$.

Alternatively, as explained in Joffan's answer, $ADBX$ is a parallelogram, so $|AD|=|XB| \; \to \; f = 3rf \; \to \; r = \frac{1}{3}$.
A: Since $AD \parallel CX$ and $BD \parallel XY$, we see $ADBX$ is a parallelogram and $|AD| = |XB|$.
Then $ |XC| = |XB|+|BC|=4\cdot |AD|$ and since $\triangle AYD \sim \triangle XYC$ we have $|XY| = 4\cdot |AY|$ and thus $|AY| : | AX| = 1:3$
