WildTrig27: Stewart's theorem

[benthurston27]

I see now, in Rational Trigonometry supplementary angles are equal, it's easy to see if you look on the spread protractor. I think it shows that all the spreads of a parallelogram are equal?

[njwildberger]

@benthurston27 That's right.

[benthurston27]

I meant in your diagram at the bottom of the screen in 1:12 you have the spread between P3 and P2 marked as r as well as between P1 and P3, and then in the proof both of those formulas have an (1-r) term instead of (1-r) and (1-r2) if P3 wasn't perpendicular to D. I have no doubt you could still prove it in the general case I was just pointing that out.

[benthurston27]

At 1:11 it looks like the original Stewart's Theorem doesn't assume that P3 meets the bottom of the triangle perpendicularly.

[njwildberger]

@benthurston27 That is correct, neither does the modern formulation I give.