SciTour: The joy of joymetry-3

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For comparison, here’s another proof. It’s equally famous, and it’s perhaps the simplest proof that avoids using areas.( Routine text book stuff)
As before, consider a right triangle with sides of length a and b and hypotenuse of length c, as shown below on the left.

Now, by divine inspiration or a stroke of genius, something tells us to draw a line segment perpendicular to the hypotenuse and down to the opposite corner, as shown above on the right. (I use to wonder why during my school days, is it just for the sake of proving it or there is no remote chance of proving it with out the line?)
This clever little construction creates two smaller triangles inside the original one. It’s easy to prove that all these triangles are “similar” — which means they have identical shapes but different sizes. That in turn implies that the lengths of their corresponding parts have the same proportions, which translates into the following set of equations:

We also know that

because our construction merely split the original hypotenuse of length c into two smaller sides of length d and e.
At this point you might be feeling a bit lost, or at least unsure of what to do next. There’s a morass of equations above, and we’re trying to whittle them down to deduce that

Nevertheless, by manipulating the right three equations, you can get the theorem to pop out. See the notes below for the missing steps.
Would you agree with me that, on aesthetic grounds, this proof is inferior to the first one? For one thing, it drags near the end. And who invited all that algebra to the party? This is supposed to be a geometry event.
But a more serious defect is the proof’s murkiness. By the time you’re done slogging through it, you might believe the theorem (grudgingly), but you still might not see why it’s true.

Reference :
E. Maor, The Pythagorean Theorem: A 4,000-Year History (Princeton University Press, 2007).
New york times.

• Here are the missing steps in the second proof above. Take this equation: