Midsegment in Triangle (Urgent!)

[kasd]

The points A1 and B1 are taken on the sides CA and CB of ABC, respectively, so that CA1 = (1 / n )* CA and CB1 = (1 / n)*CB. Prove that lines A1B1 and AB are parallel.
I have proved it but my solution isn't on a high level so pls help me!
Thanks in advance

 

[lev888]

The points A1 and B1 are taken on the sides CA and CB of ABC, respectively, so that CA1 = (1 / n )* CA and CB1 = (1 / n)*CB. Prove that lines A1B1 and AB are parallel.
I have proved it but my solution isn't on a high level so pls help me!
Thanks in advance

Please post your solution.

 

[kasd]

CA1=1/n*AC; CB1=1/n*BC
Let A1B1=x
Lets take: A2∈AC and B2∈BC such that A1A2=1/n*AC and B1B2=1/n*BC
△CA2B2: A2B2=2*A1B1=2x (midsegment)
Lets take: A3∈AC and B3∈BC such that A2A3=1/n*AC and B2B3=1/n*BC
⏢ A1A3B3B1: A2B2 - midsegment => A3B3=2*A2B2-A1B1=3x
.
.
.
We found out that: AnBn=n*x=n*A1B1 where An

A and Bn

B
Since all these lines (A1B1....) are midsegments, A1B1||A2B2||A3B3||.......||(AnBn

AB)
=> A1B1||AB
*****It's like induction, but I'm looking for sth more academical

 

[lev888]

I don't quite follow. A diagram would help.
If you are using math induction, where are the 3 steps?
I would use similar triangles.

 

[kasd]

i'm an idiot. Similar triangles would do the job. thanks btw