Browse · harp

smc

geometry intermediate

Problem

Angle

B

of triangle

A B C

is trisected by

B D

and

B E

which meet

A C

at

D

and

E

respectively. Then:

(A)

\frac{A D}{E C} = \frac{A E}{D C}

(B)

\frac{A D}{E C} = \frac{A B}{B C}

(C)

\frac{A D}{E C} = \frac{B D}{B E}

(D)

\frac{A D}{E C} = \frac{( A B ) ( B D )}{( B E ) ( B C )}

Solution

Using the Angle Bisector Theorem on

∠ A B E

and

∠ D B C

, we find that

\frac{A D}{D E} = \frac{A B}{B E}

and

\frac{D E}{E C} = \frac{B D}{B C}

. Rewriting this to better fit our answer choices gives

A D = \frac{( D E ) ( A B )}{B E}

and

E C = \frac{( D E ) ( B C )}{B D}

. Hence,

\frac{A D}{E C} = \frac{( D E ) ( A B )}{B E} \cdot \frac{B D}{( D E ) ( B C )} ⟹ \frac{A D}{E C} = \frac{( A B ) ( B D )}{( B E ) ( B C )}

.

Final answer

D