perplexus.info :: Geometry : Mystery Figures

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

Submitted by Richard

Rating: 3.5000 (2 votes)

Solution: (Hide)

The theorem of Pythagorus. The perimeter of the first figure encloses a square of side c and contains 4 right triangles congruent to ABC plus a square of side |a-b| for a total area of c² = 4×(ab/2) + (a-b)² = a²+b².
The triangle ABC of the second figure consists of the two triangles ACF and CFB which are each similar to triangle ABC. Thus area(ABC)=kc², area(ACF)=kb², and area(CFB)=ka², with the same positive value of k for each. Adding areas and canceling k again gives the result c²=a²+b². (This proof was supposedly an original creation of Albert Einstein when he was 11 years old.) A different proof, one that did not use the concept of area, was given in the comment of tomarken.

Comments: ( You must be logged in to post comments.)

	Subject	Author	Date
	Solution	tomarken	2006-03-07 07:03:31

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