Trapezoid=Parallelogram! What?

I simply love mathematical fallacies. They are a great source of mathematical entertainment, they're fun to think about and I love the looks on my friends' faces when they are stuck thinking about a mathematical fallacy.

On this post we're going to 'prove' that all trapezoids are actually parallelograms.

So, what is a trapezoid? A trapezoid is a quadrilateral [a polygon with exactly $4$ sides] with at least one pair of parallel sides. And a parallelogram is a quadrilateral with two pairs of parallel sides. Notice that according to this definition all parallelograms are trapezoids but all trapezoids are not parallelograms. Even though we know this, we're going to 'prove' otherwise. You know, just for fun!

As a quick aside you should know that what we've just called a trapezoid is called a trapezium outside of North America. This is really confusing because a in North America, a trapezium means something else [a quadrilateral with no parallel sides]. We don't want to get lost in a sea of definitions. So we're going to stick to the trapezoid-definition from the previous paragraph.

Let's begin!

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Let's call $ABCD$ a trapezoid with $AD || BC$. $BC$ is extended to $E$ such that $CE=AD$. Similarly $DA$ is extended to $F$ such that $AF=BC$.

I'm feeling a little lazy. So, I'm going to write $FA=BC=b$ and $AD=CE=a$. Also, $x=AG$, $y=GH$ and $z=CH$.

By now you should start to notice that there are some pairs of similar triangles here.

$\triangle CBG$ and $\triangle ADG$ are similar.

So, $\frac{BC}{AD}=\frac{CG}{AG}$

$\Rightarrow \frac{b}{a}=\frac{y+z}{x}$

So, $x=\frac{a(y+z)}{b} \cdots (1)$

Triangles $AFH$ and $CEH$ are similar. That means,

$\frac{AF}{CE}=\frac{AH}{CH}$

$\frac{b}{a}=\frac{x+y}{z}$.

So, $z=\frac{a(x+y)}{b} \cdots (2)$

Now we're going to subtract $(1)$ from $(2)$.

We're left with:

$z-x=\frac{a(x-z)}{b}$

$\Rightarrow |(z-x)|=|\frac{a(x-z)}{b}|$

$\Rightarrow |(z-x)|=|\frac{a}{b}|\times |(z-x)|$

$\Rightarrow |\frac{a}{b}|=1$

$\Rightarrow|a|=|b|$.

In other words, $AD=BC$. You know what this means, right? This means $ABCD$ is in fact a parallelogram. And that means all trapezoids are parallelograms.

So, what went wrong? I advise you stop reading now and think about this for a while. The solution is very simple and it's literally right under your noses [is this a correct usage of the word 'literally'?]. I also warn you that if you choose to read on you might feel like, "Darn! That was easy! I could have figured it out ."

Don't say you haven't been warned!

Congratulations to those who have figured it out or at least thought about it for a while. For those who were reading on, read on!

So, what is the fishiest step in the entire proof? The absolute value operator sure came out of nowhere! It just appeared with no rhyme or reason. When you see something fishy, that something fishy has a purpose. Here, the absolute value operator is disguising something. Let's do the steps without taking the absolute value operator. What do we get? We get this:

$\frac{a}{b}=-1$

Okay, this is weird. Because $a$ and $b$ are the lengths of the bases of the trapezoid and by convention, they're positive. So their ratio has to be positive. We have this negative ratio because we divided both of the sides by $(z-x)$. When do we see weird stuff happening after a division? When the divisor is $0$! [this not zero factorial, I'm just excited!]

$(z-x)$ is indeed $0$ and that's why when we divided the equation by it, we were left with a negative ratio that was always supposed to be positive.

Division by zero certainly doesn't destroy the space-time continuum, but it does turn all trapezoids into parallelograms!

So, $(z-x)$ is really zero? You're going to take my word for it? Right after I just presented a fallacious proof? Prove to yourself that $(z-x)$ really does equal zero or in other words, $z=x$. Post your proof in the comments section if you feel like it.

Source: I read this on Cut The Knot. This is a really great site for geometry.