Can somebody help me solve this problem about similar triangles?

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Surreal

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On side BC of △ABC point A1 is taken so that BA1 : A1C = 2 : 1. What is theratio in which median CC1 divides segment AA1?
 
Surreal said:
On side BC of △ABC point A1 is taken so that BA1 : A1C = 2 : 1. What is theratio in which median CC1 divides segment AA1?
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What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33
 
I'm sincerely sorry that i didn't obey the forum's rules

I tried solving it with the surface area but it didn't work.
P.S:Maybe if you can't help me (because of the forum's rules) you can present me wih a link to similar triangle courses.
Also it is not part of my homework(in my home country we haven't even started school yet.)
 
Surreal said:
BA1 : A1C = 2 : 1

What is the ratio in which median CC1 divides segment AA1?
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The ratio BA1 : A1C is already set.

So is the ratio AC1 : C1B because C1 is the midpoint of AB.

What ratio are you asking about? :?
 
I think i found the answer

3:1. I used similar trinagles and medicentre.
 
Surreal said:
Let point O be the point of intersection. What is the ratio AO:OA1?
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Thanks. I'm not sure why I got confused, yesterday, by the wording in your OP. Seems clear today. :oops:
 
Last edited:
Surreal said:
I think i found the answer

3:1
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I got a different ratio: 3/61*sqrt(5185) : 1

That's about 3.54 to 1 (rounded).

I picked an arbitrary triangle, and I used coordinate geometry to find the lengths of AO and OA1.

A (1, 0)

B (2, 3)

C (0, 0)

A1 (2/3, 1)

C1 (3/2, 3/2)

O (9/11, 9/11)


:idea: Maybe you forgot to specify that ∆ABC is a certain kind of triangle.
 
I came back because I had some time to experiment further.

Using an equilateral triangle, I obtained your result. :cool:

A (2, 0)

B (1, √3)

C (0, 0)

A1 (1/3, 1/√3)

C1 (3/2, √3/2)

O (3/4, √3/4)


Perhaps, the ratio is 3 : 1 for some other types of triangles, too!
 
mmm4444bot said:
I came back because I had some time to experiment further.

Using an equilateral triangle, I obtained your result. :cool:

A (2, 0)

B (1, √3)

C (0, 0)

A1 (1/3, 1/√3)

C1 (3/2, √3/2)

O (3/4, √3/4)


Perhaps, the ratio is 3 : 1 for some other types of triangles, too!
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I used medicenter and similar triangles. Maybe I have made a mistake, I am 7th grade after all!
P.S.: Where could I find geat geomtrty courses/ buy a book?
 
Surreal said:
I used medicenter and similar triangles. Maybe I have made a mistake …
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Maybe, but we can't say, unless you post your work.


… Where could I find [great geometry] courses … [books]?
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I have not reviewed anything, but you could certainly start by googling. There are lot's of good resources available on-line.

For videos, you could try keywords:

khan academy geometry videos

For textbooks, you could try keywords:

free online geometry textbooks

For written courses, you could try keywords:

free online geometry courses

free online geometry exercises worksheets

Do not provide personal or payment information, at any site, and skip the ads. 8-)
 
PS: If you decide to show your work, be sure to also include the original exercise statement, including all given information.

Also, please read the forum guidelines, if you have not already done so. Cheers :cool:
 
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