, 2026

Xander's rope is $10 \text{ m}$ long.

Since Yasmin's rope is $n$% longer
than Xander's rope, then the length of Yasmin's rope is $10\left(1 + \dfrac{n}{100}\right)\text{ m}$.

Since Zhe's rope is $(2n)$% longer
than Yasmin's rope, then the length of Zhe's rope is $10\left(1 + \dfrac{n}{100}\right)\left(1 + \dfrac{2n}{100}\right)\text{ m}$.

Since Zhe's rope is $(3.14n)$% longer
than Xander's rope, then the length of Zhe's rope can also be written as
$10\left(1 + \dfrac{3.14n}{100}\right)\text{ m}$.

Therefore, $$\begin{align*} 10\left(1 + \dfrac{n}{100}\right)\left(1 + \dfrac{2n}{100}\right) & = 10\left(1 + \dfrac{3.14n}{100}\right)\\ \left(1 + \dfrac{n}{100}\right)\left(1 + \dfrac{2n}{100}\right) & = \left(1 + \dfrac{3.14n}{100}\right)\\ (100 + n)(100 + 2n)& = 100(100 + 3.14n) & \text{(multiplying by $100 \cdot 100$)}\\ 10000 + 300n + 2n^2 & = 10000 + 314n \\ 2n^2 - 14n & = 0\\ 2n(n-7) & = 0\end{align*}$$ Since $\text{n gt; 0}$, then it must be the case that
$n = 7$.
Solution 1:

Let $BC = x$ and $CD = y$. Join $A$ to $C$.

Using the cosine law in $\triangle ABC$, we obtain $$\begin{align*} AC^2 & = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) \\ & = 16 + x^2 - 8x\cos(45\degree) \\ & = 16 + x^2 - 8x\!\left(\frac{1}{\sqrt{2}}\right) \\ & = 16 + x^2 - 4\sqrt{2}x\end{align*}$$ Using the cosine law
in $\triangle ADC$, we obtain $$\begin{align*} AC^2 & = AD^2 + DC^2 - 2(AD)(DC)\cos(\angle ADC) \\ & = 16 + y^2 - 8y\cos(135\degree) \\ & = 16 + y^2 - 8y\!\left(-\frac{1}{\sqrt{2}}\right) \\ & = 16 + y^2 + 4\sqrt{2}y\end{align*}$$ Equating expressions
for $AC^2$, we obtain $$\begin{align*} 16 + x^2 - 4\sqrt{2}x & = 16 + y^2 + 4\sqrt{2}y \\ x^2 - y^2 - 4\sqrt{2}x - 4\sqrt{2}y & = 0\\ (x+y)(x-y) - 4\sqrt{2}(x+y) & = 0\\ (x+y)(x-y - 4\sqrt{2}) & = 0\end{align*}$$ Since $\text{x gt; 0}$ and $\text{y gt; 0}$, then $\text{x + y gt; 0}$. Thus, $x - y - 4\sqrt{2} = 0$ and so $BC - CD = x - y = 4\sqrt{2}$.

Solution 2:

Let point $P$ be on $BC$ so that $AP$ is perpendicular to $AB$.

To see why $P$ is on $BC$ (and not some extension of $BC$) first observe that isosceles $\triangle BAD$ has $\text{ADB = ABD lt; ABC = 45}$, so $$\text{BAD = 180 - ADB - ABD gt; 180 - 45 - 45 = 90}$$ Therefore, $\angle BAD$ is obtuse.

Now suppose $P$ were on some
extension of $BC$. Since $\angle BAD$ is obtuse and $\angle BAP=90\degree$, $AP$ must intersect $CD$ at some point $M$, and so $\text{AM lt;AP}$. However, $AP=AB=4$ since $\triangle BAP$ is is a right-isosceles
triangle, which means in $\triangle AMD$, we have that $AM$ is
not the longest side while it is opposite obtuse $\angle ADM$. This is impossible, so we
conclude that $P$ must be on $BC$.

It was mentioned above that $\triangle BAP$ is right-angled and isosceles, with $AP = AB = 4$ which means that $BP = \sqrt{2}AB = 4\sqrt{2}$.

Since $\angle BPA = 45\degree$ and
$BPC$ is a straight angle, then $\angle CPA = 180\degree - \angle BPA = 135\degree$. Therefore, $\angle APC = \angle ADC$.

Since $AP=AD=4$, then $\triangle APD$ is isosceles, and so $\angle APD = \angle ADP$. Then $$\angle CPD = \angle APC - \angle APD = \angle ADC - \angle ADP = \angle CDP$$ Since $\angle CPD = \angle CDP$, then $\triangle CPD$ is isosceles, and so $CD = CP$. Thus, $BC - CD = BC - CP = BP = 4\sqrt{2}$.