The Student Room Group
Reply 1
please help me.... T.T i have exams tommrow
Reply 2
argh..............
Reply 3
which exam is that? All of them ??? and you need help at this time ??
Ok, I'll try the 1st
Reply 4
thank you so much!!! i have my high school math exam tommorow!!! and it's only 9pm cuz its canada
Reply 5
Do you know about the meeting point of median lines in a triangle ???

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My suggestion is:
You take X' is the midpoint of CD, then you prove that AX' cuts BD at O' which statisfies: BO' = O'D.

-> O' = O => X' = X.

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I need to go to sleep btw It's very very late here. Sorry
Reply 6
i have to use vector method.... there is going to be number of vector proof question on the exam
Reply 7
jeongsuman
i have to use vector method.... there is going to be number of vector proof question on the exam

Ah, ok, then let a = AB, b = AD
I am sure you can find the expression of AO in term of a and b

Then you may find CX and DX in tems of a and b
You'll see CX = XD
Reply 8
how????? -_-
Reply 9
jeongsuman
how????? -_-

It seems too late now. Yesterday, I had to go to bed, that time was 3am already, and I had lecture at 9am. Sorry.

Let AB = a, AD = b

=>AC = a + b

So AO = 1/3 a + 2/3b

Let |DX|/|DC| = k

=> 1-k = |XC|/|DC|

So AX = kAC + (1-k)AD = k(a +b) + (1-k)b

=> AX = ka + b

Since AX has the same direction with AO

Then compare the coefficients of a and b we have
k÷(1/3) = 1÷(2/3)
Then you'll get k = 1/2

So DX = XC = 1/2DC
Reply 10
the same question appeared on the exam and i couldn't solve it... anyway thanks for solving the question
Reply 11
So did you do well with other questions??
Reply 12
yes other questions were pretty straightforward