D is midpoint of side BC …

Given:-In fig., {tex}\Delta{/tex}ABC D is the mid-point of BC and E is the mid-point of AD.

To prove:- BE : EX = 3 : 1
Construction: Through D, draw DF parallel to BX.
Proof:- In {tex}\Delta{/tex}AEX and {tex}\Delta{/tex}ADF
{tex}\angle EAX = \angle DAF{/tex} [Common]
{tex}\angle AXE = \angle AFD{/tex} [Corresponding angles]
Then, {tex}\Delta{/tex}AEX ~ {tex}\Delta{/tex}ADF [By AA similarity]
{tex}\therefore \frac{{EX}}{{DF}} = \frac{{AE}}{{AD}}{/tex} [Corresponding parts of similar {tex}Triangle{/tex}are proportional]
{tex}\Rightarrow \frac{{EX}}{{DF}} = \frac{{AE}}{{2AE}}{/tex} [AE = ED given]
{tex}\Rightarrow{/tex} DF = 2EX ....(i)
In {tex}\Delta{/tex}CDF ~ {tex}\Delta{/tex}CBX
{tex}\angle C = \angle C{/tex} [Common]
{tex}\angle CDF = \angle CBX{/tex} [Corresponding angles]
Then, {tex}\Delta{/tex}CDF ~ {tex}\Delta{/tex}CBX [By AA similarity] 

Therefore ,{tex} \frac{{CD}}{{CB}} = \frac{{DF}}{{BX}}{/tex} [Corresponding parts of similar {tex}\Delta{/tex}are proportional]
{tex}\Rightarrow \frac{1}{2} = \frac{{DF}}{{BE + EX}}{/tex} [BD = DC given]
{tex}\Rightarrow{/tex} BE + EX = 2DF
{tex}\Rightarrow{/tex} BE + EX = 4EX [By using (i)]
{tex}\Rightarrow{/tex} BE = 4EX - EX
{tex}\Rightarrow{/tex} BE = 3EX{tex}{/tex}
{tex}\Rightarrow \frac{{BE}}{{EX}} = \frac{3}{1}.{/tex}