Question

In the given triangle ABC, D is mid-point of AB. P is on AC such that $PC=\frac{1}{2}AP$ and $DE||BP$. If $AE=xAC$, then find the value of $x$.

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is B $\frac{1}{3}$

In $△ABP$,

D is the mid-point of AB and $DE||BP$
$\therefore E$ is mid-point of $AP$.
$\therefore AE=EP$
also,
$PC=\frac{1}{2}AP$
$\Rightarrow 2PC=AP$
$\Rightarrow 2PC=2AE$
$\Rightarrow PC=AE$
$\therefore AE=PE=PC$
$\Rightarrow AC=AE+EP+PC$
$\Rightarrow AC=AE+AE+AE$
$\Rightarrow AC=3AE$
$\Rightarrow AE=\frac{1}{3}AC$
Therefore,
$x=\frac{1}{3}$