Thread: 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


Adam

Hey adam23.

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.

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?

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.

I can use the vectors but how can I do it?

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).

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?

Mayby you know, how can I use this, that the ARPC and BRQC are a half of hexagon?

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.

I have no idea how I should do this. Chiro if you can please write me the solution

You may want to take your teachers advice and your own suggestion about the hexagon by shifting BRQC where R is at point A and you will get that hexagon you were talking about.