If $x$ is proportional to $y$ and $x$ is proportional to $z$, then is $x \propto yz$ or $x^2 \propto yz$? Suppose $x$ is proportional to $y$ and $x$ is proportional to $z$. Then we can write,
\begin{align*}
  x &= k_{1}y
\end{align*}
The constant $k_{1}$ is going to depend on $z$. So,
$$k_{1} = f(z)$$
Since $x$ is proportional to $z$,
$$k_{1} = f(z) = cz$$, for some constant $c$.
So we have,
$$x = czy $$
$$x \propto zy $$
But, since $x$ is proportional to $z$ we can write,
$$x = k_{2}z$$
So multiplying the two equations we have,
$$x^{2} = k_{1}k_{2}yz$$
$$\frac{x^{2}}{yz} = k_{1}k_{2}$$
$$x^{2} \propto yz$$
These are two conflicting results. So what am I missing here? I have looked at this thread: How does one combine proportionality?
and while it does give some insight, doesn't fully answer my question.
Edit: Ok, I think I understand it now. In the last part we had $x^2 = k_1k_2yz$, since we are considering $y$ and $z$ as things that can change, we must also consider the constants. We know that if $z$ changes, $k_1$ changes, and if $y$ changes, $k_2$ changes. That is, $k_1 = c_1z$ and $k_2 = c_2y$. So we have
\begin{align}
x^2 &= c_1c_2y^2z^2 \\
x^2 &\propto (yz)^2 \\
\end{align}
From this can I write
$$x \propto (yz)$$ ?
 A: The two results are both correct under different interpretations of what "$x$ is proportional to $y$ and $x$ is proportional to $z$" means.
One degree of freedom
One interpretation is that $x$, $y$, and $z$ are all related in a way that knowing any one of them determines the other two. Geometrically, the set of all points $(x,y,z)$ permitted by the relationship between them is a curve in space.
In this case, "$x$ is proportional to $y$" gives us a relationship $x = k_1 y$ that is absolute: if I tell you what $y$ is, you can immediately use $x = k_1y$ to find $x$. The same thing applies to the statement "$x$ is proportional to $z$": we have $x = k_2 z$, with some different constant.
In this case, $x^2 = k_1 k_2 yz$ is true but slightly misleading: it might make you think that you can choose any value of $y$ and $z$ and use the formula to find what $x^2$ is. In fact, from $x = k_1y$ and $x = k_2z$ we can determine that $z = \frac{k_1}{k_2} y$: if we choose a value of $y$, that determines both $x$ and $z$.
Still, it's true that $x^2$ is proportional to $yz$, in the sense that the ratio between $x^2$ and $yz$ is always the same.
Two degrees of freedom
Another interpretation is that $y$ and $z$ are separate inputs, independently free to take on any value we like. Geometrically, the set of points $(x,y,z)$ permitted by the relationship between them is a surface in space.
In this case, "$x$ is proportional to $y$" should really be said as "$x$ is proportional to $y$ when $z$ is held constant". We have $x = k_1 y$, but $k_1$ is a value that depends on $z$.
If $x$ is proportional to $y$ when $z$ is held constant, and $x$ is proportional to $z$ when $y$ is held constant, then:

*

*Because $\frac{x}{y} = k_1$, where $k_1$ does not depend on $y$, the quantity $\frac{x}{yz} = \frac{k_1}{z}$ also does not depend on $y$.

*Because $\frac{x}{z} = k_2$, where $k_2$ does not depend on $z$, the quantity $\frac{x}{yz} = \frac{k_2}{y}$ also does not depend on $z$.

We conclude that $\frac{x}{yz}$ does not depend on either $y$ or $z$: it is a true constant. Therefore $x$ is proportional to the product $yz$.
A: Explanation

Suppose x is proportional to y and x is proportional to z.

Mathematically, this means that $x=k_1y$ and $x=k_2z$, where $k_1$ and $k_2$ are constants. Multiply these two equations together: $x^2=k_1k_2yz$. Let $k_3=k_1k_2$. $x^2=k_3yz$, and since $k_3$ is a constant ($k_2$ and $k_1$ are constants), this means that $x^2\propto yz$.
Where you're going wrong
Like @David_Quinn mentioned, you are going wrong when you say that the constant $k_1$ is going to depend on $z$, or when you say that "we know that if $z$ changes, $k_1$ changes, and if $y$ changes, $k_2$ changes". Constants are constants; they don't change. The fact that $x$ is proportional to $z$ means that for any $x$, $z$ is equal to that $x$ divided by a constant $k_1$, and that constant has to be the same regardless of $x$ and $z$.
Testing with numbers
We can also arrive at the same conclusion by testing out numbers, although this is hardly rigorous. Let $x=1$, $y=2$, and $z=3$. This means that $k_1=2$ and $k_2=3$. If $x$ were to change to $2$, then, $z$ would change to $6$ and $y$ would change to $4$. $yz$ is $6$ when $x$ is $1$ and $24$ when $x$ is $2$. $x^2$ is $1$ when $x$ is $1$ and $4$ when $x$ is $2$. Since $\dfrac{24}{6}=4$, we can say that $yz$ might be proportional to $x^2$, but cannot be proportional to $x$. This is the same as what we get from our analysis above.
Edit
The other answer is talking about a case where $z\propto k_1$, and not $z\propto x$. This might be why you got confused and wrote $k_1=f(z)$. In your particular case, however, this is not true, so $A$ is not proportional to $BC$.
