## Sunday, November 18, 2007

### Circles, Chords, Tangents, Similar Triangles and that Ubiquitous 3-4-5 Triangle

[As always, don't forget to give proper attribution when using the following in the classroom or elsewhere as indicated in the sidebar]

The cone in the sphere problem led me to an interesting relationship in the corresponding 2-dimensional case with a surprise ending. (Only a math person would compare a math problem to a mystery novel!). The following investigation allows the student to explore a myriad of possibilities: from similar triangles to the altitude on hypotenuse theorems to Pythagorean, to chord-chord or secant-tangent power theorems, coordinate methods, draw the radius technique, etc. Sounds like this one problem might review over 50% of a geometry course? You decide for yourself! Just remember -- one person is not likely to think of every method. Open this up to student discovery and watch miracles unfold...

In the diagram above, segment AF is a diameter of the circle whose center is O, BC is a tangent segment (F is the point of tangency), BC = AF and BF = FC. Segments AB and AC intersect the circle at D and E, respectively. Lots of given there! Perhaps some unnecessary information?

(a) If AF = 40, show that DE = 32.
Notes: To encourage depth of reasoning, consider requiring teams of students to find at least two methods.

(b) Let's generalize (of course!). This time no numerical values are given. Everything else is the same. Prove, in general, that DE/BC = 4/5.

(c) So where's the 3-4-5 triangle (one similar to it, that is)? Find it and prove that it is indeed similar to a 3-4-5.

Totally_Clueless said...

Very nice, Dave. This is pretty cool.

As an aside, it took me a longer time to prove that DG=GE, and that angle AGD was right, than to do the rest of the problem assuming these were true.

TC

Dave Marain said...

tc--
You are absolutely correct that it is non-trivial to prove what most students would 'assume' is true! From my perspective the key is point F, although you may have found an alternative approach. We do not know that A is equidistant from D and E but we can demonstrate (with some effort) that F is! I'll omit the details for now so that others can play around with it. Note that I did give a hint by mentioning 'altitude on hypotenuse' theorems! Since O is equidistant from D and E it follows that the line containing O and F is the perpendicular bisector of DE!

I'm glad you appreciated this subtlety. I think this would prove fairly challenging and you're correct that the rest of the problem is easy to complete. Remember, however, I am suggesting to the student that there is a myriad of approaches here and they need to come up with more than one method in their group. This is one way to stretch the stronger student who believes they have solved the problem in rapid fashion and they just sit there complacently.

Kaz Maslanka said...

Very nice!

K