2018 AMC 12A Problem 18

Problem 18 of 25IntermediateGeometry

Triangle

ABC

with

AB = 50

and

AC = 10

has area

120.

Let

D

be the midpoint of

\overline{AB},

and let

E

be the midpoint of

\overline{AC}.

The angle bisector of

\angle BAC

intersects

\overline{DE}

and

\overline{BC}

at

F

and

G,

respectively. What is the area of quadrilateral

FDBG?

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Solution

Since

D

and

E

are midpoints,

\triangle ADE

has

\tfrac14

the area of

\triangle ABC,

namely

30,

so trapezoid

EDBC

has area

120 - 30 = 90.

By the Angle Bisector Theorem,

G

divides

BC

with

BG = \tfrac{AB}{AB + AC} \cdot BC = \tfrac56 BC,

and likewise

F

divides

DE

so that

DF = \tfrac56 DE.

Because

FDBG

and

EDBC

share the same height, the area of

FDBG

is

\tfrac56

of the area of

EDBC:

\tfrac56 \cdot 90 = 75.

Thus, the correct answer is D.

More practice

Concepts: angle bisector theorem · area ratio · midpoint

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.