You have to prove that BE:EX =3:1

3 years, 1 month ago

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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

- 3 years, 1 month ago

Nice solution :).

- 3 years, 1 month ago

- 3 years, 1 month ago

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

- 3 years, 1 month ago

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

- 3 years, 1 month ago

can we do it by applying mid point theorem?

- 3 years, 1 month ago

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

- 3 years, 1 month ago

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

I think Actually we can prove this by 3 ways.

- 3 years, 1 month ago

I am glad you have found your answers. As always in math, there are many ways to do things.

- 3 years, 1 month ago

Yeah thank you.

- 3 years, 1 month ago

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

- 3 years, 1 month ago