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Geometry Difficulty 5.0 AIME, harder Find the answer

1811018 \cdot 110 As shown, ABCDA B C D is an isosceles trapezoid, ADA D =BC=5,AB=4,DC=10=B C=5, A B=4, D C=10, point CC is on DFD F, and BB is the midpoint of the hypotenuse of the right triangle DEF\triangle D E F. Then CFC F equals
(A) 3.25 .
(B) 3.5 .
(C) 3.75 .
(D) 4.0 .
(E) 4.25 .

Multiple choice: answer with the letter of the option you want.

Solution

[Solution] Draw BHDFB H \perp D F, then BH//EFB H / / E F. Since ABCDA B C D is an isosceles trapezoid,
CH=12(DCAB)=3, \therefore \quad C H=\frac{1}{2}(D C-A B)=3,

then
DH=DCCH=103=7 D H=D C-C H=10-3=7 \text {. }
Since BB is the midpoint of DED E, and DBDE=DHDF\frac{D B}{D E}=\frac{D H}{D F},
DF=14 \therefore D F=14 \text {. }

Then CF=DFCD=4C F=D F-C D=4.
Therefore, the answer is (D)(D).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.