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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 $. 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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