1. ## Difficult Proof

Given that $A*B = C*D$, prove that $\frac{D – B}{\frac{B + D}{2}} = \frac{C – A}{\frac{A + C}{2}}$.

Attempted solution:

$A*B = C*D$

$\Rightarrow \frac{B}{D} = \frac{C}{A}$

$\Rightarrow 1 - \frac{B}{D} = 1 - \frac{C}{A}$

$\Rightarrow \frac{D}{D} - \frac{B}{D} = \frac{A}{A} - \frac{C}{A}$

$\Rightarrow \frac{D - B}{D} = \frac{A - C}{A}$ ....

Where do I go from here?

2. ## Re: Difficult Proof

Originally Posted by RogueDemon
Given that $A*B = C*D$, prove that $\frac{D – B}{\frac{B + D}{2}} = \frac{C – A}{\frac{A + C}{2}}$.
This is false, for example, for A = 3, B = 4, C = 2, D = 6. However, the conclusion is true if AD = BC, B + D ≠ 0 and A + C ≠ 0.

3. ## Re: Difficult Proof

Would the conclusion be true if (D - B) and (C - A) were absolute values? If so, would it be possible to prove this?

4. ## Re: Difficult Proof

Interesting. Yes, if AB = CD and neither A = C = 0 nor B = D = 0, then $\left|\frac{B-D}{B+D}\right|=\left|\frac{A-C}{A+C}\right|$. Under these assumptions, B = 0 iff C = 0 and A = 0 iff D = 0. Check what happens when B = 0 and C = 0, or when A = 0 and D = 0. The remaining case is when neither of the numbers equals 0. Then AB = CD implies A / C = D / B. Let's call this number x. We have

$\left|\frac{B-D}{B+D}\right|=\left|\frac{A-C}{A+C}\right|$ iff

$\left|\frac{1-D/B}{1+D/B}\right|=\left|\frac{1-C/A}{1+C/A}\right|$ iff

$\left|\frac{1-x}{1+x}\right|=\left|\frac{1-1/x}{1+1/x}\right|$ iff

$\left|\frac{1-x}{1+x}\right|=\left|\frac{x-1}{x+1}\right|$

which is true.