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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Updated on: 10-Oct-2022

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