Thales' theorem and point in circle

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 $p_i,p_j,p_l$.

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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

2 Answers

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 $\beta'<\beta$.

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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.

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