24. A rectangle $P Q R S$ has side-lengths $a$ and $b$, with $a<b$. The rectangle $P T U V$ has side-lengths $c$ and $d$, with $c<d$. Also, $a<d$ and $c<b$, as shown. The sides $R S$ and $T U$ cross at $X$.
Which of these conditions guarantees that $Q, X$ and $V$ lie on a straight line?
A $\frac{a}{b}+\frac{c}{d}=1$
B $\frac{a}{c}+\frac{b}{d}=1$
C $\frac{a}{d}+\frac{c}{b}=1$
D $\frac{a}{c}+\frac{d}{b}=1$
$\mathrm{E} \frac{c}{a}+\frac{b}{d}=1$
24. C The gradient of $V Q$ is $-\frac{P V}{Q P}=-\frac{d}{b}$. The gradient of $X Q$ is $-\frac{T X}{Q T}=-\frac{a}{b-c}$. Therefore, for $Q, X$ and $V$ to lie on a straight line, it is necessary that $\frac{d}{b}=\frac{a}{b-c}$. Hence $a b=b d-c d$, which simplifies to $\frac{a b+c d}{b d}=1$. This in turn simplifies to $\frac{a}{d}+\frac{c}{b}=1$.