>Let me make sure I've got this.
>
>I don't understand "extended intersect" or "in K". I think
>what you are saying is that K is a point halfway between B
>and C and therefore along the same line as AQ if AQ was
>extended. Correct?

Yes, it is.

>How do you know that QK=AQ/3.

Triangle ABC is equilateral. The line AK is the median, the altitude, the angle (A) and the side (BC) bisector. Its being the median is important. In a sense, Q is also almost everything: barycenter (center fo gravity), circumcenter, incenter, orthocenter, etc.

The important thing is that Q is the center of gravity of the points A, B, C with equal masses, say, 1. Then K is the center of gravity of points B and C. Now, place a mass of 2 = (1 + 1) in K, and think of the center of gravity of A with mass 1 and K with mass 2. It'll be 1/3 of the way from K.

>Is that a known law or is
>there a formula that tells you that. I have the same
>question for SO=SQ*3/4.

Exactly for the same reason, right. O is the center of gravity of SABC and could be found as the center of gravity of two masses: 1 at S and 3 at Q.

>I've learned the theorem of cosines but I don't remember it
>off the top of my head. I've been known to get sine,
>cosine, cosecant and cotangent mixed up.

Just search this site or the web.