The Secant Theorem

So, if you drag point \(\small\mathtt{C}\) around in the diagram above, you will see that the two expressions \(\small\mathtt{f \cdot (d + f)}\) and \(\small\mathtt{g \cdot (e + g)}\) are always equal. This result is known as the Secant (SEE-knt) Theorem--also called the Secant Segment Theorem.

Here's how it goes. The lines above, \(\small\mathtt{\overleftrightarrow{GC}}\) and \(\small\mathtt{\overleftrightarrow{IC}}\), are called secant lines of the circle, because they are straight lines and they each intersect the circle at two points. (I haven't drawn them as extending past point \(\small\mathtt{C}\), but they are lines, so they do.) These particular secant lines meet at a point outside of the circle (point \(\small\mathtt{C}\)). And secant lines of a circle that meet outside of the circle are the kinds of secant lines that are important to the Secant Theorem.

You can see that four segments--called secant segments now--are created: two internal secant segments (\(\small\mathtt{d}\) and \(\small\mathtt {e}\)) and two external secant segments (\(\small\mathtt{f}\) and \(\small\mathtt{g}\)). The Secant Theorem says that \(\small\mathtt{f \cdot (d + f) = g \cdot (e + g)}\).



Part of the Why

Download Chrome here. It's like 15 seconds, and it's way better.

When you move your mouse over the diagram at the right, you should see two triangles: \(\small\Delta\)\(\small\mathtt{CAB}\) and \(\small\Delta \)\(\small\mathtt{CED}\). The two \(\small\mathtt{\alpha}\) symbols show congruent angles. These angles are congruent because they are inscribed angles that both mark off the same arc on the circle. (Since they're each half the measure of the [invisible] central angle that marks off that green arc, they have to be equal.) The \(\small\mathtt{\beta}\) symbol shows that both triangles share \(\small\mathtt{\measuredangle{C}}\).

According to the AA Similarity Theorem, the two triangles are similar, which means that the ratios of the lengths of their corresponding sides are equal. Click on the circle a few times to see the corresponding sides. We can see that the equation below is true:
\[\mathtt{\frac{m\overline{CB}}{m\overline{CD} + m\overline{CA}} = \frac{m\overline{CD}}{m\overline{CB} + m\overline{BE}}}\]

And "we" (meaning "I") have to click and look back at that equation every time to convince myself that it's true because of similarity. Anyway, this means that after a little cross-multiplying action, we've got \(\small\mathtt{m\overline{CB} \cdot (m\overline{CB} + m\overline{BE}) = m\overline{CD} \cdot (m\overline{CD} + m\overline{CA})}\), which is a restatement of the Secant Theorem.



Some Questions We Didn't Ask

This is another situation where I drew auxiliary lines to demonstrate the proof. But could I always draw auxiliary lines for this proof? I was able to create similar triangles in the diagram just above, and it looks like I could do it for every position I could get to in the GeoGebra applet at the top. But not every possibility is represented there.

Also, we didn't talk about what corresponding really means (and a lot of textbooks that cover geometry kind of breeze past it too). The standard definition of corresponding angles is something like "angles that are in the same relative position." Which is ambiguous to say the least.