Problem:
Let ABCD be an isosceles trapezoid with AD=BC and AB<CD. Suppose that the distances from A to the lines BC,CD, and BD are 15,18, and 10, respectively. Let K be the area of ABCD. Find 2​⋅K.
Solution:
By symmetry, â–³ABC is congruent to â–³BAD. The areas of these two triangles can be calculated in three ways as
215⋅BC​=218⋅AB​=210⋅BD​.
Thus there is a constant k such that AB=5k,AD=BC=6k, and AC=BD=9k. Because ABCD is an isosceles trapezoid, it is cyclic, so by Ptolemy's Theorem, AC⋅BD=AB⋅CD+AD⋅BC. Thus CD=9k. Let P be the foot of the perpendicular from B to CD. Then CP=2CD−AB​=2k. By the Pythagorean Theorem, BC2=BP2+CP2, which implies that (6k)2=182+(2k)2 and k=492​​. Therefore the area of ABCD is
