1952 AMC 12 Problem 19

Question

Angle

B

of triangle

ABC

is trisected by

BD

and

BE,

which meet

AC

at

D

and

E,

respectively. Then:

\dfrac{AD}{EC}=\dfrac{AE}{DC}

\dfrac{AD}{EC}=\dfrac{AB}{BC}

\dfrac{AD}{EC}=\dfrac{BD}{BE}

\dfrac{AD}{EC}=\dfrac{AB\cdot BD}{BE\cdot BC}

\dfrac{AD}{EC}=\dfrac{AE\cdot BD}{DC\cdot BE}

Compare the areas of triangles

ABD

and

EBC

in two different ways

They have bases on

AC

, and their included angles at

B

are equal trisection angles

Triangles

ABD

and

EBC

have bases

AD

and

EC

on the same line

AC,

so their common altitude from

B

gives

\frac{[ABD]}{[EBC]}=\frac{AD}{EC}.

Also,

\angle ABD=\angle EBC

because both are one-third of angle

B.

Using two sides and the included angle for the same area ratio gives

\frac{[ABD]}{[EBC]} =\frac{AB\cdot BD}{BE\cdot BC}.

Thus, the correct answer is D.

Problem 19 in other years