Question

In the adjoining figure, $AD$ and $BE$ are the medians of $△ABC$ and $DF\parallel BE$. Show that $CF=\frac{1}{4}AC$.

Answer 1

Given:

In $\Delta ABC$,

$AD$ and $BE$ are the medians.

$DF\parallel BE$

To prove: $CF=\frac{1}{4}AC$

Proof:

Consider $\Delta ABC$

Since $AD$ is Median $⟹BD=DC\cdots \left(1\right)$

Since $BE$ is Median $⟹AE=EC\cdots \left(2\right)$

Consider $\Delta BEC$

Since $BE||DF$

By Basic Proportionality theorm,

$⟹\frac{BD}{DC}=\frac{EF}{FC}$

$⟹\frac{DC}{DC}=\frac{EF}{FC}$ [Since, $BD=DC$]

$⟹1=\frac{EF}{FC}$

$⟹CF=EF$ ---(3)

Now, from (2),

$AE=EC$

$⟹AE=2EF$

$⟹EF=\frac{1}{2}AE$

$⟹EF=\frac{1}{2}\left(\frac{AC}{2}\right)$ [Since, $E$ is midpoint of $AC$.]

$⟹EF=\frac{AC}{4}$ ----(4)

Now, from (3),

$CF=EF$

$⟹CF=\frac{1}{4}AC$

Hence, proved.