### Forming a proof about a triangle

Posted:

**Thu Dec 02, 2010 1:03 am UTC**Hi. I'm trying to prove something, and I don't really know if this form is considered valid.

Suppose you have a right triangle, with hypotenuse z and the other two legs designated as x and y. Using the fact that the area A=(z^2)/4 , I want to prove that x=y, aka that the triangle is isosceles. I will also end up using the fact that the area A=(1/2)xy and the Pythagorean theorem, z^2 = x^2 + y^2.

My problem is that in my proof, i take the identity (1/2)xy=(z^2)/4, and do a handful of operations to either side of it, which eventually transform it to x=y. I am a first year undergraduate, and don't have much experience with proofs, but I'm taking a course on discrete math that's supposed to serve as an introduction to proofs, among other things, and all the direct proofs I've seen, and the explanation in my text, are such that you take one expression (not an equality) and transform it to another. So in this case, I would take x, and transform it to y. Im having trouble finding a way to do this, so is what i have something that is considered correct? (btw this is not homework, it's just me wandering through problems in the text, for my own edification)

1. (z^2)/4 = (1/2)xy

2. Z^2 = 2xy (multiply both sides by 4)

3. x^2 + y^2 = 2xy (substitute by the pythagorean theorem)

4. (x^2 + y^2)/(xy) = 2 (multiply both sides by xy)

5. (y/x) + (x/y) = 2 (decompose the fraction on the left side)

6. Y/x = 2 - x/y (subtract x/y from both sides)

7. Y = 2x - (x^2)/y (multiply both sides by x)

8. Y = x(2 - x/y) (factor right side)

9. y^2 = x(2y - x) (multiply both sundes by y)

10. (d/dx)(y^2) = (d/dx)[x(2y - x)] (multiply both sides by d/dx)

11. 2y(dy/dx) = x[2(dy/dx) - 1] + (2y - x) (carry out derivetives)

12. 2y(dy/dx) = 2x(dy/dx) - x - 2y - x (distribute on right side)

13. 2y(dy/dx) - 2y = 2x(dy/dx) -2x (subtract 2y from both sides)

14. 2y[(dy/dx) - 1] = 2x[(dy/dx) - 1] (factor both sides)

15. 2y = 2x (divide both sides by (dy/dx) - 1 )

16. y = x (guess)

Suppose you have a right triangle, with hypotenuse z and the other two legs designated as x and y. Using the fact that the area A=(z^2)/4 , I want to prove that x=y, aka that the triangle is isosceles. I will also end up using the fact that the area A=(1/2)xy and the Pythagorean theorem, z^2 = x^2 + y^2.

My problem is that in my proof, i take the identity (1/2)xy=(z^2)/4, and do a handful of operations to either side of it, which eventually transform it to x=y. I am a first year undergraduate, and don't have much experience with proofs, but I'm taking a course on discrete math that's supposed to serve as an introduction to proofs, among other things, and all the direct proofs I've seen, and the explanation in my text, are such that you take one expression (not an equality) and transform it to another. So in this case, I would take x, and transform it to y. Im having trouble finding a way to do this, so is what i have something that is considered correct? (btw this is not homework, it's just me wandering through problems in the text, for my own edification)

1. (z^2)/4 = (1/2)xy

2. Z^2 = 2xy (multiply both sides by 4)

3. x^2 + y^2 = 2xy (substitute by the pythagorean theorem)

4. (x^2 + y^2)/(xy) = 2 (multiply both sides by xy)

5. (y/x) + (x/y) = 2 (decompose the fraction on the left side)

6. Y/x = 2 - x/y (subtract x/y from both sides)

7. Y = 2x - (x^2)/y (multiply both sides by x)

8. Y = x(2 - x/y) (factor right side)

9. y^2 = x(2y - x) (multiply both sundes by y)

10. (d/dx)(y^2) = (d/dx)[x(2y - x)] (multiply both sides by d/dx)

11. 2y(dy/dx) = x[2(dy/dx) - 1] + (2y - x) (carry out derivetives)

12. 2y(dy/dx) = 2x(dy/dx) - x - 2y - x (distribute on right side)

13. 2y(dy/dx) - 2y = 2x(dy/dx) -2x (subtract 2y from both sides)

14. 2y[(dy/dx) - 1] = 2x[(dy/dx) - 1] (factor both sides)

15. 2y = 2x (divide both sides by (dy/dx) - 1 )

16. y = x (guess)