Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone explain to me why $\alpha' < \alpha$ and $\beta'<\beta$ when point $p_l$ is inside the circle? There is suppose to be a way to see this using Thales' theorem.

Also if $p_l$ is inside the circle defined by $p_i,p_j,p_k$ then how does one show that $p_k$ is inside the circle defined by $p_i,p_j,p_l$.

enter image description here

share|cite|improve this question
In your picture it looks like $p_l$ is actually the center of the circle, which can be somewhat misleading. – tomasz Jul 29 '12 at 11:04
Also, thoughts? Especially about the second one – I can't see what can be hard about this one, so I can't really provide a hint. Or do you mean by "in the circle" actually "within the disc bounded by the circle"? That is obviously false (like in your picture, if you swap $p_k$ with either $p_j$ or $p_i$). – tomasz Jul 29 '12 at 11:07
up vote 0 down vote accepted

For a chord on a circle, say $p_ip_k$, the angle subtended from a point, say $p_j$ is half the angle subtended from the center; so $\alpha'<\alpha$. A similar argument for $\beta'<\beta$.

share|cite|improve this answer
I had the impression that the question was about a point inside the circle, not in the center. And you did not really use Thales here (directly). :) – tomasz Jul 30 '12 at 11:33

Looks like homework, so I'll try to just drop some hints.

For the first one, replace $p_j$ with a $p_j'$, the intersection of the line $\overline{p_ip_l}$ and the circle. Then consider the line through $p_l$ parallel to $\overline{p_j'p_k}$ and its intersection with $\overline{p_ip_k}$.

For the second one, notice that for arbitrary circle, any three points on it define the very circle, so it doesn't matter which three you pick.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.