Just construct DF parallel to BX . So by MPT in BXC. DF = 1/2 BX. Now DF // EX. So by MPT in triangle ADF. We get XE = 1/2 DF. i.e. XE = 1/4 BX and so BE:EX = 3:1

@Maria Kozlowska
–
draw a line parallel to BC through the point E. say PQ,then in triangle ABD, E is mid point of AD and also,EP parallel to BD,therefore AP=BP,and similarly AQ=QC.Now Triangle BXC is similar to Triangle EXQ,
Therefore,BX/EX=XC/XQ=BC/EQ.Let us take,BX/EX=BC/EQ,i.e BX/EX=BC/1/2DC,
i.e BX/EX=BC/1/2(1/2B/C),That gives,BX/EX=4/1,
Now subtracting 1 from both sides,that gives,
BE:EX=3:1

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestJust construct DF parallel to BX . So by MPT in BXC. DF = 1/2 BX. Now DF // EX. So by MPT in triangle ADF. We get XE = 1/2 DF. i.e. XE = 1/4 BX and so BE:EX = 3:1

Log in to reply

Nice solution :).

Log in to reply

You have Given that BD=CD and AE=DE.

Log in to reply

If you apply Menelaus' theorem twice you will get that ratio.

Log in to reply

can we do it by applying mid point theorem?

Log in to reply

I do not see how it could be helpful in this case.

Log in to reply

BE:EX=3:1I think Actually we can prove this by 3 ways.Log in to reply

Log in to reply

Log in to reply

Log in to reply

yeah but we can also do it by the use of mid point theorem , just draw a line parallel to BC through point E.and further you can solve

Log in to reply