# mathematics

ABCD is a rectangle, in which BC=2AB. A point E lies on ray CD, such that CE=2BC. Prove that BE perpindicular AC

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1. 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)
so
angle BFC = 90 degrees

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2. thanks for ur help..

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3. Thanks

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4. Thank you damon

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