posted by math... on .
ABCD is a rectangle, in which BC=2AB. A point E lies on ray CD, such that CE=2BC. Prove that BE perpindicular AC
let AB = x
then BC = 2x
then CE = 4x
angle B and angle C are right
AB/BC = BC/CE = 1/2
triangle ABC similar to triangle BCE
Call F intersection of BE and AC
Now angle EBC = FAB (similar triasngles)
Angle FCB = BEC (also same similar triangles)
BUT those angles add to 90 degrees (two angles in right triangle)
angle BFC = 90 degrees
thanks for ur help..