Can you find BC?

First notice that dc-ab=8 which is equal to the base of the triangle that you can make by drawing an imaginary line on the left of the trapezium. Then use the Pythagorean theorem by doing C^2-A^2=B^2.. 289-64=225 square root of 225 is 15.

Using the pythagorean theorem, we have

$BC=\sqrt{17^2-8^2}=\sqrt{225}=$ $\boxed{15}$

we construct a line parallel to AD from vertex B onto side CD to meet at E. so BE=17...NOW in triangle BEC angle C=90. so by pythagoras theorem we get BC=15 as EC=8 and BE=17

if we draw a perpendicular from A to the line CD ( which cuts the line CD at point , say E ) the length of the perpendicular will be the altitude of the triangle ADE as well as the altitude of the trapezium............ now we can say that AECB is a rectangle and hence DE= CE = 8.......
applying Pythagoras theorem in the triangle ADE we get the length of BC ( which is equal to AE) as 15