# Triangle

• Oct 29th 2012, 07:42 AM
Triangle
Hello!

I have a problem with this one:

Points P, Q, R lie respectively on sides BC, CA and AB of ABC triangle. AR=RP=PC and BR=RQ=QC. Prove that AC + BC =2 AB

Thanks for help in advance

• Oct 29th 2012, 07:58 PM
chiro
Re: Triangle

Can you draw a picture (or do you have one)?

This is one of the best techniques to solving problems (not just mathematical by the way, but problems in general) and this is the first thing you should do any of these sorts of geometry problems.
• Oct 29th 2012, 10:55 PM
Re: Triangle
• Oct 30th 2012, 12:00 PM
Re: Triangle
My math's teacher said that I should something extra draw to this picture but I don't have any idea. Mayby you know what?
• Oct 30th 2012, 06:18 PM
chiro
Re: Triangle
There is a way you can do a simple proof with vectors, but I will ask if this is OK with you. I understand that in high school, vectors aren't taught that often so if this is a class exercise (and I assume it is), then you will need to use other techniques.

If you can't use the vector approach, you should outline the kinds of techniques you are covering at the moment.
• Oct 31st 2012, 07:14 AM
Re: Triangle
I can use the vectors but how can I do it?
• Oct 31st 2012, 04:41 PM
chiro
Re: Triangle
You have AB + BC = AR + RP + PC and BA + AC = BR + RQ + QC as vectors (but not as lengths). Also BA = -AB since reversing directions reverses the vector.

Now add these together and consider the lengths of the vector (AB + BA + BC + AC = AB - AB + BC + AC = BC + AC).
• Nov 2nd 2012, 10:43 AM
Re: Triangle
Ok, I got that vectors AR+RP+PC+BR+RQ+QC=BC+AC, but I still don't got AC+BC=AB. How can I consider the lengths of the vector?
• Nov 3rd 2012, 12:24 AM
Re: Triangle
Mayby you know, how can I use this, that the ARPC and BRQC are a half of hexagon?
• Nov 3rd 2012, 04:27 AM
chiro
Re: Triangle
Show us what you got vector wise and what you reduced it down to: you won't get a direct vector answer but rather will have to use identities relating to sine's and cosines that help show the lengths are the same.

Remember that you get lengths of vectors with inner and outer products where <a,b> = ||a||*||b||*cos(a,b) and a X b = n*||a||*||b||*sin(a,b) where is a normal vector of unit length perpendicular to a and b.
• Nov 4th 2012, 03:19 AM