2018 AMC 12A Problem 18

Question

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?

60

65

70

75

80

Answer: D

Concepts:angle bisector theorem area ratio midpoint

Difficulty rating: 1990

Small hint:

Since

D, E

are midpoints,

[ADE] = \tfrac14 [ABC],

so trapezoid

EDBC

is the remaining

\tfrac34

Big hint:

By the Angle Bisector Theorem the bisector cuts

DE

and

BC

in ratio

AB : AC = 5 : 1,

making

FDBG

equal to

\tfrac56

of

EDBC

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.