As shown in the figure , $ABCD$ is a trapezoid，$AD\parallel BC$ , $\angle C=45^{\circ}$, $AB=AD=4$ ，$E$ is a point on the line $DA$ , $EF\perp BE$ and $AB\perp AD$.
Show: $BE=EF$.
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What do you mean by "half line"? – DonAntonio May 01 '14 at 19:54
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@DonAntonio I believe he means the ray $DA$. – David H May 01 '14 at 20:01
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Is F a point on the line DC? – nadia-liza May 01 '14 at 20:01
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@nadia-liza Yes. – Railgun May 01 '14 at 20:02
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1@Railgun The segment $AB$ appears perpendicular to $AD$ by looking at the diagram, so, just to clarify (in case you forgot to mention it), is $AB\perp AD$ ? – user26486 May 01 '14 at 20:04
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@DavidH, that sounds logical. Thanks. – DonAntonio May 01 '14 at 20:04
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You forgot to mention the fact that $AB\perp AD$. I've created a diagram on GeoGebra without making the angle between $AB$ and $AD$ perpendicular and I got that $BE$ was far from being equal to $EF$. When I made it perpendicular, the equality was satisfied. I'll edit the question. – user26486 May 01 '14 at 20:26
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First as mathh suggested, you need $AD \perp BC$ (which is part of the conclusion), or else your conclusion need not actually be true. This is because you can construct a diagram with your given conditions without the extra condition; however, $BE = EF$ also implies $AD \perp BC$ given your diagram.
Since we know $\triangle BEF$ is a right triangle, it suffices to show $\angle EBF = 45^{\circ}$.
Draw the line $BD$, then $\triangle ABD$ is a right triangle, and since it is isosceles we know $\angle ABD = 45^{\circ}$, so $\angle DBC = 45^{\circ}$ as well. Then $\triangle DBC$ is a right triangle as well. Since the opposite angles of the quadrilateral $BEFD$ add to $180^{\circ}$ we know it is cyclic.
So $\angle EBF = \angle EDF$ which is equal to $\angle C = 45^{\circ}$ (since $AD$ and $BC$ are parallel), and we are done.
Kevin
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