a vector problem

chandra21

chandra21

Junior Member
Joined
May 23, 2013
Messages
66
friends,i am sloving piliminary vector problems.but can't solve one of them.here it goes,

Q : ABCD is a parallelogram and E is the mid-point of CD and F is a point on AE such that AF=(2/3)AE ;show that F lies on the diagonal BD and BF=(2/3)BD.
 
please direct me.

i try it in this way- first i have consider the triangle ADC.Since E is the mid point of DC,then AE is a median of the triangle.since F divides AE in the ratio 2 : 1 ,then the centroid of the triangle at F. If the diagonals of the parallelogram meet at P then P is the midpoint of AC. then DP is a median of the triangle which passes through the centroid F. So F is on the diagonal BD. How can i show that
BF=(2/3)BD
 
i am new here.so i am not familiar with this site. If someone atleast reply i will very much thankful to him.
 
friends,i am sloving piliminary
vector problems.but can't solve
one of them.here it goes,
Q : ABCD is a parallelogram and E
is the mid-point of CD and F is a
point on AE such that AF=
(2/3)AE ;show that F lies on the
diagonal BD and BF=(2/3)BD.
i try it in this way- first i have consider the triangle ADC.Since E is the mid point of DC,then AE is a median of the triangle.since F divides AE in the ratio 2 : 1 ,then the centroid of the triangle at F. If the diagonals of the parallelogram meet at P then P is the midpoint of AC. then DP is a median of the triangle which passes through the centroid F. So F is on the diagonal BD. How can i show that
BF=(2/3)BD .please help
 
chandra21 said:
friends,i am sloving piliminary
vector problems.but can't solve
one of them.here it goes,
Q : ABCD is a parallelogram and E
is the mid-point of CD and F is a
point on AE such that AF=
(2/3)AE ;show that F lies on the
diagonal BD and BF=(2/3)BD.
i try it in this way- first i have consider the triangle ADC.Since E is the mid point of DC,then AE is a median of the triangle.since F divides AE in the ratio 2 : 1 ,then the centroid of the triangle at F. If the diagonals of the parallelogram meet at P then P is the midpoint of AC. then DP is a median of the triangle which passes through the centroid F. So F is on the diagonal BD. How can i show that
BF=(2/3)BD .please help
Click to expand...

I think you are making this too difficult. Consider, you have two similar triangles: BFA and DFE. Corresponding sides of similar triangles will all have the same ratio.
 
chandra21 said:
is there anyone for help?
Click to expand...
This is not a staffed chat room with people waiting, on-call, to provide instant replies. This is a public forum where helpers are volunteers with families, jobs, and schedules; they surf by when they're able. Posters should expect to wait hours (at least!) for a reply.

Also, when you reply so many times to yourself, the listing for your post looks like an ongoing conversation, so many (most?) helpers will assume that you're already getting help from somebody else.

With respect to your first reply (to yourself), the statements are unclear. For instance, "then the centroid of the triangle at F" has no verb, so it is unclear what you are intending to say. Kindly please reply with clarification. Thank you! ;)
 
Given: ABCD is a parallelogram.
By definition, AB is parallel to CD.
AB and BD are transverse lines cutting across parallel lines AB and CD.
They form "alternate interior angles", which are congruent:
e.g., angle ABD is congruent to angle EDB
and angle BAE is congruent to angle DEA.
It can also be seen that angle BFA is congruent to angle DFE, since these are "vertical angles".

There are many examples of similar triangles available on the Internet. Here are some:

http://library.thinkquest.org/20991/geo/spoly.html

http://www.mathsisfun.com/geometry/triangles-similar.html
 
it is clear to me. But you have already assume that F lies on BD,which is to be proved? Pls let me know it.
 
i will keep it in my mind. I am a new member and log in today. I am Sorry for that. And tha
will be 'the centroid of the triangle is at F'. Is it always required a perfect grammatical sentence?
 
chandra21 said:
it is clear to me. But you have already assume that F lies on BD,which is to be proved? Pls let me know it.
Click to expand...

I did not know if you recognized the similar triangles since you had not stated it. I was hoping you would continue on from there. I should not have called the point of intersection F yet, as you correctly point out. I just wanted to make sure you saw which angles and triangles I was talking about. Call it some other point initially if you wish, say Z.

After establishing the similar triangles BZA and DZE, note that DE and BA are corresponding sides. Since E is a midpoint of DC, DE is (1/2) of DC and is therefore also (1/2) of BA. This establishes the ratio between ALL corresponding sides of the triangles.

We can now say that BZ is twice the length of ZD. This also means that BZ is (2/3) the length of BD.

Since it is given that "F lies on the diagonal BD and BF=(2/3)BD", point F and point Z must be the same point.
 
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