Side Intercept

Geometry Level 4

In the triangle $ABC$ , point $D$ is the midpoint of $AB$ . Point $E$ is on side $AC$ , such that $BE$ intersects $CD$ at point $F$ , with $| DF | = 2 | FC|$ . What is the ratio $\dfrac{|CE|}{|AC|}$ ?

0.2

\frac{1}{3}

0.4

0.25

3 solutions

Swapnil Das
Aug 26, 2016

Construction : Construct $DG\parallel BE$ .

Now, $\large AG=GE$ by midpoint theorem, say equal to $\large 2y$ .

Again, in $\large \triangle DGC$ , $\large \frac { CF }{ FD } =\frac { CE }{ EG }$ by Basic Proportionality Theorem, which implies $\large EC=y$ .

$\huge \therefore \frac { CE }{ AC } =\frac { y }{ 2y+2y+y } =\frac { 1 }{ 5 }$ .

Ahmad Saad
Aug 16, 2016

Niranjan Khanderia
Aug 20, 2016

Applying the formula derived from the fig. on the left we solve the fig to the right.

Side Intercept

3 solutions

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