21. The diagram shows a triangle $X Y Z$. The sides $X Y, Y Z$ and $X Z$ have lengths 2,3 and 4 respectively. The lines $A M B, P M Q$ and $S M T$ are drawn parallel to the sides of triangle $X Y Z$ so that $A P, Q S$ and $B T$ are of equal length.
What is the length of $A P$ ?

A $\frac{10}{11}$
B $\frac{11}{12}$
C $\frac{12}{13}$
D $\frac{13}{14}$
$\mathrm{E} \frac{14}{15}$
SOLUTION
C
Let the common length of $A P, Q S$ and $B T$ be $x$.
Because $A B$ is parallel to $X Z$, the corresponding angles, $\angle P A M$ and $\angle Y X Z$ are equal. Because $P Q$ is parallel to $Y Z$, the corresponding angles, $\angle A P M$ and $\angle X Y Z$ are equal.

It follows that the triangles $A P M$ and $X Y Z$ are similar. Therefore their corresponding sides are in the same ratio. Hence, in particular, $\frac{A P}{X Y}=\frac{A M}{X Z}$. That is, $\frac{x}{2}=\frac{A M}{4}$, and therefore $A M=2 x$. Since $A M$ is parallel to $X S$, and $X A$ is parallel to $S M$, the quadrilateral $M A X S$ is a parallelogram. Therefore $X S=A M=2 x$.
Similarly, the triangles $M T B$ and $X Y Z$ are similar, and therefore $\frac{T B}{Y Z}=\frac{M B}{X Z}$. Hence $\frac{x}{3}=\frac{M B}{4}$. Therefore $M B=\frac{4}{3} x$. Also $M B Z Q$ is a parallelogram. Therefore $Q Z=M P=\frac{4}{3} x$.
Now, as $X S+S Q+Q Z=X Z$, it follows that
$$
2 x+x+\frac{4}{3} x=4 .
$$

That is,
$$
\frac{13}{3} x=4
$$
from which it follows that
$$
x=\frac{3}{13} \times 4=\frac{12}{13} \text {. }
$$

Therefore the length of $A P$ is $\frac{12}{13}$.
FOR INVESTIGATION
21.1 Find the lengths of $X A$ and $P Y$ in terms of $x$. Then use the fact that $X Y$ has length 2 to determine the value of $x$.
21.2 Similarly, determine the value of $x$ by finding the lengths of $Y T$ and $B Z$, and then use the fact that $Y Z$ has length 3 .
21.3 Is it true that for every triangle $X Y Z$ there is a point $M$, such that when the lines $A M B$, $P M Q$ and $S M T$ are drawn parallel to the sides of the triangle, then $A P, Q S$ and $B T$ are of equal length ?