D is the mid-point of side BC of a $\triangle ABC$. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that $BE:EX=3:1$.


Given:
D is the mid-point of side BC of a \( \triangle A B C \). AD is bisected at the point E and BE produced cuts AC at the point \( X \).
To do:
We have to prove that $BE:EX=3: 1$.
Solution:
Construction: Draw DY parallel to BX.
In $\vartriangle ADY$, using midpoint theorem
$EX=\frac{DY}{2}$
In $\vartriangle BCX$, using midpoint theorem
$DY=\frac{BX}{2}$
Therefore,
$EX=\frac{BX}{4}$ 
$4EX=BX$
$4EX=BE+EX$
$3EX=BE$
$\frac{BE}{EX}=\frac{3}{1}$
$BE:EX=3:1$
Hence proved.

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Updated on: 10-Oct-2022

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