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If $\dfrac{a}{b}=\dfrac{c}{d}$, show that $\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{(a+2c)(a+3c)}{(b+2d)(b+3d)}$

$\dfrac{a}{b}=\dfrac{c}{d}=k$

$\Rightarrow a=bk, c=dk$

$\dfrac{(bk)^2-(dk)^2}{b^2-d^2}=\dfrac{(bk+2dk)(bk+3dk)}{(b+2d)(b+3d)}$

$[(bk)^2-(dk)^2][(b+2d)(b+3d)]=(b^2-d^2)[(bk+2dk)(bk+3dk)]$

$k^2(b^2-d^2)(b^2+5bd+6d^2)=(b^2-d^2)(k^2)(b+2d)(b+3d)$

$k^2(b^2-d^2)(b+2d)(b+3d)=(b^2-d^2)(k^2)(b+2d)(b+3d)$

$1=1$

It looks correct to me, but I'm a bit confused what the end result $1=1$ means since it looks so trivial and pointless. Does it mean I've proven $\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{(a+2c)(a+3c)}{(b+2d)(b+3d)}$? Thanks.

Bill Dubuque
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ronald christenkkson
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    No, you must start from this: $\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{(a+2c)(a+3c)}{(b+2d)(b+3d)}$ – Lục Trường Phát Aug 26 '23 at 06:49
  • By the dupe $,k=\color{#c00}{\dfrac{a}b}=\color{#0a0}{\dfrac{c}d},\Rightarrow, k = \dfrac{m:!\color{#c00}a+n:!\color{#0a0}c}{m:!\color{#c00}b:!+n:!\color{#0a0}d},$ so clearly both sides $=k^2$ (after factoring LHS). – Bill Dubuque Oct 01 '23 at 07:07

2 Answers2

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Yes, your proof is correct but after the first step, following second step should quickly finish the problem : $$k^2\frac{b^2-b^2}{b^2-d^2}=k^2\frac{(b+2d)(b+3d)}{(b+2d)(b+3d)}\Rightarrow k^2 = k^2$$ For an insight, it is well known $$\text{If} \quad \frac{a}{b}=\frac{c}{d}=k \quad \text{then} \quad \frac{a}{b}=\frac{c}{d}=\frac{ax+cy}{bx+dy}=k$$ for real numbers $x,y\,$ such that $bx+dy \neq 0$. Putting $(x,y)=(1,1), (1,-1), (1,2), (1,3)$ in order, we see that $$\frac{a}{b}=\frac{c}{d}=k=\frac{a+c}{b+d}=\frac{a-c}{b-d}=\frac{a+2c}{b+2d}=\frac{a+3c}{b+3d}$$ The four ratios involved in question are all equal (to $k$) and so is product of first and last two (to $k^2$).

MyMolecules
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You've got the right general idea. However, your calculations are more complicated than need be, and you also should be careful about using what you're trying to prove in your solution. Nonetheless, apart from where $k = 0$, since all of your operations are otherwise reversible, you have basically got your proof.

Here's a method that is somewhat shorter, and which doesn't assume what's requested to be proven. Using your substitution of $\frac{a}{b}=\frac{c}{d}=k$, as you've already indicated, the LHS of equality to prove is

$$\frac{a^2-c^2}{b^2-d^2} = \frac{(bk)^2-(dk)^2}{b^2-d^2} = \frac{k^2(b^2-d^2)}{b^2-d^2} = k^2\left(\frac{b^2-d^2}{b^2-d^2}\right) = k^2$$

and, similarly after the substitution, the RHS becomes

$$\frac{(bk+2dk)(bk+3dk)}{(b+2d)(b+3d)} = \frac{k(b+2d)k(b+3d)}{(b+2d)(b+3d)} = k^2\left(\frac{(b+2d)(b+3d)}{(b+2d)(b+3d)}\right) = k^2$$

Thus the $2$ sides are equal, proving the equality. However, one small thing to note is that I believe the question implicitly has that $b \ne \pm d$ since, otherwise, with $b = \pm d$ we also have $a = \pm c$, so the LHS then becomes the indeterminate $\frac{0}{0}$. Similarly, there's also an assumption that $b + 2d \neq 0$ and $b + 3d \neq 0$, since either one being $0$ means the RHS then becomes $\frac{0}{0}$.

John Omielan
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