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 Mystery Figures (Posted on 2006-03-07)
The following figure pertains to the proof of what theorem and why?

Here ABC, B'AE, A'B'C', and BA'D are congruent right triangles with corresponding vertices each written in the same order, C is on the line AE, E is on the line B'C', and C' is on the line A'D.

```
A

C    D     B

B'    E    C'

A'

```

Explain how the same theorem has a proof that utilizes the following figure.

Here ABC is a right triangle and CF is it's altitude.

```       C

A  F         B ```

 See The Solution Submitted by Richard Rating: 3.5000 (2 votes)

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 Solution Comment 1 of 1

I believe they both represent proofs of the Pythagorean Theorem.  It's been a long time since I've done any geometry, but I'll give this a shot:

In the first figure, the four triangles create a large square (ABA'B') with a smaller square "hole" in the middle (CDC'E).  To make matters simpler, let's call the shortest side of each triangle a, the longer side b, and the hypotenuse c.  The area of each triangle is ab/2.  The area of the large square is c^2, and the area of the smaller square is (b-a)^2.  From that we get:

c^2 = 4(ab/2) + (b-a)^2

c^2 = 2ab + b^2 - 2ab + a^2

c^2 = b^2 + a^2

I'm not sure about the second figure, but I'm assuming that all three triangles are similar (ABC, AFC, and CFB).  That gives you the following ratios:

AC/AB = AF/AC   or   AC^2 = AB * AF

BC/AB = BF/BC   or   BC^2 = AB * BF

Then, AC^2 + BC^2 = AB(AF + BF) = AB^2

Hope I didn't make a mistake there. :)

 Posted by tomarken on 2006-03-07 07:03:31

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