Hello everyone!

Welcome back to Pythagoras Week.  If you missed it, be sure to check it out from the beginning.  Our final proof doesn't rely on any sort of fancy results like the last few.  We just need a few copies of our triangle.

Take three copies of a right triangle with sides .  Scale one by , one by , and the third by .  Place them together as shown below.The angles work out nicely, and it turns out we get a rectangle.  The top edge has length and the bottom edge has length .  Opposite sides of a rectangle are the same length, so we get .

That's it for Pythagoras week.  I'll post a final wrap-up tomorrow.

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Hello everyone!

Welcome back to Pythagoras Week.  If you missed it, be sure to check it out from the beginning.  Our proof today makes use of similar triangles.  We call two triangles "similar" if they have all the same angles, as in the left diagram below.  Actually, because the angles in a triangle add to , we only need to know that two angles are the same to show similarity, as on the right.The really nice feature of similar triangles is that corresponding sides are all in the same proportions.  So if triangles and are similar, then .

Once more, let's take a right triangle with sides , as below.  Let's also add a perpendicular line from to , as shown, breaking the side into pieces of length and .Now we've got two smaller triangles, and .  Each has a right angle and shares an angle with , so they both have two angles in common with our original triangle.  As we said above, this means they're similar to each other and to .

As I mentioned earlier, corresponding sides are proportional.  So .  Rewriting this in terms of side lengths, we get , and .  Similarly (no pun intended), we get .  Thus, .

Join me tomorrow for one more proof.  I promise I've chosen a nice one for last.

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Hello everyone!

Welcome back to Pythagoras Week.  If you missed it, be sure to check it out from the beginning.  Today, we'll use a result known as "power of a point".  Take a circle with center and a point .  Draw a line through that is tangent to the circle at point and another which meets the circle at two points, and , as shown below.  Under these conditions, .The full statement of power of a point is a bit more general, but this will work for today.  We'll use this fact without proof, but if you're curious, a full proof can be found here.

Now let's use power of a point to prove the Pythagorean Theorem.  We'll start with our right triangle .  Draw a circle around which passes through .  Extend to , as shown.Then power of a point gives us .  Writing all these lengths in terms of , we get .  A little algebra gives us .

That's it for today.  Tune in tomorrow for another proof.

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