Pythagorean Theorem Proof #20?

Can you explain to me step by step how this works? D: It really looks like As Bs and Cs mixed up all together. I used this site:

I'll copy and paste it here too. [except for the image]

"This one is a cross between #7 and #19. Construct triangles ABC', BCA', and ACB' similar to ABC, as in the diagram. By construction, ΔABC = ΔA'BC. In addition, triangles ABB' and ABC' are also equal. Thus we conclude that Area(A'BC) + Area(AB'C) = Area(ABC'). From the similarity of triangles we get as before B'C = AC2/BC and BC' = AC·AB/BC. Putting it all together yields AC·BC + (AC2/BC)·AC = AB·(AC·AB/BC) which is the same as

BC2 + AC2 = AB2."

1 Answer

  • 1 decade ago
    Best Answer


    First, the diagram is made by drawing triangles that are similar to the original right triangle ABC. All three of the new triangles are in proportion to the original. It is important that you understand what similar triangles are before you attempt to go further with this proof.

    Next, we want to recall that in a right triangle with an altitude to the hypotenuse (a line from the right angle to the long side that crosses the long side at a right angle), the altitude divides the original right triangle into two similar right triangles. Furthermore, the length of the altitude is geometric mean of the two parts the hypotenuse was split into.

    So if you have right triangle ABC with B the right angle and AC the hypotenuse, draw BD where BD is perpendicular to AC. What you will find is that:

    AD / BD = BD / DC


    AD = (BD)²/DC

    This is the concept that he is trying to get at when he writes:

    "From the similarity of triangles we get as before B'C = AC²/BC and BC' = AC·AB/BC."

    Now, what he says next took a while to grasp, but now, it seems elementary.

    The way the added triangle were constructed, the AREA of triangle AB'C + area of triangle ABC = area of triangle ABC'

    If that isn't apparent, just draw it yourself, cut it out and rearrange the pieces.

    So the next line:

    "Putting it all together yields AC·BC + (AC2/BC)·AC = AB·(AC·AB/BC)"

    Is just saying the same thing.

    (Don't forget that the areas of triangles have that 1/2 part to it, but since all three areas come from triangles, he just factored it out.)

    Basically, he wrote Base*Height for three different triangles:

    AC*B'C + AC*BC = AB*BC'

    And just substituted the first equations to eliminate B'C and BC'.

    From there, it is just algebra.

    Hope this clears it up. If not, Email me.

Still have questions? Get your answers by asking now.