Prove that if the two angles on the base of a trapezoid are different, then the diagonal starting from the smaller angle is longer than the other diagonal.
Solution
∠ABC < ∠DCB iff A'B > CD'. Put AA' = h, A'D' = a, then BD2 = h2 + (a + A'B)2, AC2 = h2 + (a + CD')2, so BD > AC iff A'B > CD'.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03