Remote Learning III: Secant-Secant

(Click on the comic if you can't see the full image.)

(C)Copyright 2020, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

What do Math teachers have? Well, for one, T-shirts like that one!

Guesses to what my actual shirt says can be left in the comments.

So, if I reteach the same material year after year, I'm allowed to re-use the same puns, right? It's all new to them.

Since I said that I would explain on my blog, I guess I need to do that here.

As we see in Panel 2, above, we are Given Secant AB with point D on the circle, and AC with point E on the circle.
We want to prove that the products of these lengths are equal: (AB)(AD) = (AC)(AE)

If we draw chords CD and BE, we create triangles ABE and ACD, as shown in Panel 3.
Angles B and C both intercept the same arc, DE, and therefore they are congruent.
Angle A is congruent to itself because of the Reflexive Property.
Therefore, triangles ABE and ACD are similar.
If they are similar, then their corresponding sides are proportional.

So AB / AE = AC / AD
If we cross-multiply, we get: (AB)(AD) = (AC)(AE)

Or, in other words, "The whole line times the other part equals the other whole line times its outside part".

Actually, not very long, and could easily be included in an actual remote video, but not so much in a four- or six-panel comic page.




Come back often for more funny math and geeky comics.

Problem: Pythagorean Theorem and Tangent-Secant

There are multiple ways to solve problems with circles, but some will be more straightforward than others, based on the information given.

If a circle has a tangent drawn to it, then that tangent is perpendicular to the radius it intersects. In other words, a right angle is created. Generally speaking, that right angle will probably signal the need to use the Pythagorean Theorem at some point in the problem.

On the other hand, if the tangent is accompanied by a secant line, then a second theorem can be plucked from our toolkit: the Tangent-Secant Theorem. If the tangent and secant intersect at a point outside the circle, then the square of the length of the tangent from the external point to the circle will be equal to the product of the portion of the secant outside the circle times the length of the entire segment.

Consider the problem below:

Which of the two theorems do we need to use?

The answer is: either one of them.

The circle has three radii drawn, but only one is labeled. Write the "6" next to the other two segments.

You can now solve for x using the right triangle with legs 6 and x, and with hypotenuse 10. Or you can solve for tangent with length x using the secant with a length of 16 and an external length of 4.

If you choose to work both of them out, you'll find the same answer.

Keeping that in mind, look at this next problem:

Now you will see that you can make a diameter from the given radius, and create a secant. This gives you a second option for some for x.