Thanks in advance to anyone who can help me out on this. I'm currently a junior in high school taking and doing well my school's honors pre-calc class, but of all of the math I've ever learned, proofs have always given me trouble. And that's what we've just begun re-learning, my kryptonite. Anyways, I just can't figure out how to prove this problem like it says.
Prove that if the diagonals of a trapezoid are congruent, then the trapezoid is isosceles, using coordinate geometry.
I'm solely restricted to using things like the midpoint formula, distance formula, slope formula, etc. I can't use any theorems from geometry other than the Pythagorean Theorem.
I've tried drawing a trapezoid with the points (0,0), (a, b), (a+c, b), and (d,0), and then since the given is that the diagonals are equal, finding the distances of the diagonals and setting them equal, and solving them for one variable. Then I took that variable and then plugged it into the distances of the two legs of the trapezoid, since an isosceles trapezoid has two congruent legs. Unfortunately, I just can't get the two legs of the trapezoid to end up being congruent. If anyone can explain how to solve this problem, that would be awesome!
Hint: Let the vertices, in counterclockwise order, be $P,Q,R,S$, with $PQ$ parallel to $RS$.
Let $P=(0,0)$, $Q=(a,0)$, $R=(b,h)$, and $S=(c,h)$.
Because the diagonals are equal, you know that $b^2+h^2=(c-a)^2+h^2$.
You want to show that the two non-parallel sides are equal, so you want to show that $c^2+h^2=(b-a)^2+h^2$.