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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

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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

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Nice solution :).

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

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

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can we do it by applying mid point theorem?

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

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

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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

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