In the figure, ABCD is an isosceles trapezoid with side lengths AD=BC=5, AB=4, and DC=10. The point C is on DF and B is the midpoint of hypotenuse DE in the right triangle DEF. Then CF=

1989 AMC 12 Problem 4

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In the figure,

ABCD

is an isosceles trapezoid with side lengths

AD=BC=5,

AB=4,

and

DC=10.

The point

C

is on

\overline{DF}

and

B

is the midpoint of hypotenuse

\overline{DE}

in the right triangle

DEF.

Then

CF=

3.25

3.5

3.75

4.0

4.25

Drop perpendiculars from

A

and

B

\overline{DF}

The segment through midpoint

B

parallel to

EF

reaches the midpoint of

DF

Drop perpendiculars

AG

and

BH

DF.

In the isosceles trapezoid,

GH=AB=4

and

DG=HC=\frac{DC-AB}{2}=3.

Since

B

is the midpoint of

DE

and

BH\parallel EF,

the midpoint theorem makes

H

the midpoint of

DF.

Hence

DH=DG+GH=7,

DF=14

and

CF=14-10=4.

Thus the correct answer is D.