Question

Let $ABC$ be a triangle and $D$ be a point on the segment $BC$ such that $DC=2BD.$ Let $E$ be the mid-point of $AC.$ Let $AD$ and $BE$ intersect in $P.$ Determine the ratios $\frac{BP}{PE}$ and $\frac{AP}{PD}$.

• A. $\frac{BP}{PE}=1$
• B. $\frac{BP}{PE}=3$
• C. $\frac{AP}{PD}=1$
• D. $\frac{AP}{PD}=3$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{BP}{PE}=1$
D $\frac{AP}{PD}=3$

From figure,
We get $CF/FD=1=CE/EA$
$\therefore EF||AD$
Hence EF || PD
$\therefore BP/PE=BD/DF$
But $BD=DF$
$\therefore BP/PE=1$
In triangle ACD, since $EF||AD$,
we get $EF/AD=CF/CD=1/2$
$thereforeAD=2EF$
But PD/EF = BD/BF = 1/2
Hence EF = 2PD
$\therefore AP=ADPD=3PD$
We getAP/PD = 3.
(Coordinate geometry proof is also possible.)