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Hexagon $ABLCDK$ is inscribed in a circle. Line $LK$ cuts line segments $AD, BC, AC, BD$ in points $M, N, P, Q$ respectively, prove $|NL||KP||MQ|=|KM||PN||LQ|$.

So, I have the solution, but I don't understand it.

The solution as given:

Let $$s=\sin\frac{AB}2,t=\sin\frac{BL}2,u=\sin\frac{LC+AK}2,\\v=\sin\frac{CK}2,w=\sin\frac{DK}2,x=\sin\frac{LD+AK}2.$$

Then we have:

$$\frac{|NL|\cdot|KP|\cdot|MQ|}{|KM|\cdot|PN|\cdot|LQ|}=\\ =\frac{|NL|}{|NC|}\cdot\frac{|NC|}{|NP|}\cdot\frac{|KP|}{|KA|}\cdot\frac{|KA|}{|KM|}\cdot\frac{|QM|}{|QD|}\cdot\frac{|QD|}{|QL|}=\\ =\frac tv\cdot\frac us\cdot\frac vu\cdot\frac xw\cdot\frac sx\cdot\frac wt = 1$$

What I don't understand is why is $\frac{|NL|}{|NC|}=\frac tv$, or why is $\frac{|NC|}{|NP|}=\frac us$. (I don't understand the others too but maybe I will if you explain those 2 to me.)

So my main question: Can you explain one or both of those 2 to me?

Second question (if not answered already): Why do we have $\sin\frac{AB}2$? Why sine of a side? What do we get from this?

Edit: So, $\sin\frac{AB}2$ refers to sine of half of the arc AB (not side). So, only the main question remains.

Edit: Question is answered in comments.

Pero
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    I believe that $\sin\frac{AB}{2}$ isn't the sine of (half of) a side, $\overline{AB}$; it's the sine of (half of) an arc, $\stackrel{\frown}{AB}$. In general, for chord $\overline{PQ}$ of a circle with diameter $d$, $$|\overline{PQ}| = d;\sin\frac{\stackrel{\frown}{PQ}}{2}$$ This principle (together with the Inscribed Angle Theorem) is typically used in demonstrating the Law of Sines. – Blue Feb 09 '19 at 14:43
  • @Blue So the best I could come up with is $\frac{\sin\frac{BL}2}{\sin\frac{CK}2}=\frac{|BL|}{|CK|}$. How do I get to $\frac{|NL|}{|NC|}$? Is it because $BL$ and $CK$ are opposite equal angles? – Pero Feb 09 '19 at 15:31
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    By the Law of Sines in $\triangle NCL$, $$\frac{|NL|}{|NC|} = \frac{\sin\angle C}{\sin\angle L} = \frac{\sin(\stackrel{\frown}{BL}/2)}{\sin(\stackrel{\frown}{CK}/2)} = \frac{t}{v}$$ – Blue Feb 09 '19 at 17:53
  • @Blue Oh, I see. One more question: can you explain why $\frac{|NC|}{|NP|} = \frac{\sin\frac{LC+AK}2}{\sin\frac{AB}2}$ and not $\frac{|NC|}{|NP|} = \frac{\sin\frac{LC}2}{\sin\frac{AB}2}$? Why is it $\sin\frac{LC+AK}2$ instead of $\sin\frac{LC}2$? – Pero Feb 09 '19 at 18:33
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    $\angle NPC$ is not an inscribed angle, so it's not half of a particular arc ... The edges of that angle determine two arcs: $\stackrel{\frown}{CL}$ and $\stackrel{\frown}{AK}$. The only fair thing to do (which happens to be the correct thing to do) is average those arcs: the angle measure is half the sum of the arcs. (Note that the Inscribed Angle Theorem actually does this: it averages the subtended arc with a "zero" arc. Moreover: If lines through a circle meet at a point outside the circle, then the angle measure is half the difference of the subtended arcs.) – Blue Feb 09 '19 at 18:39
  • Okay, I understand now. Thank you very much! – Pero Feb 09 '19 at 18:43
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    This discussion would make more sense to people with a picture. I'll post one as an answer shortly. – Blue Feb 09 '19 at 18:46

1 Answers1

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In this figure ...

enter image description here

... we can apply the Law of Sines in the two shaded triangles, then invoke the Inscribed Angle Theorem and a relative:

$$\begin{align} \triangle CNL: &\quad\frac{|NL|}{|NC|}= \frac{\sin\angle BCL}{\sin\angle CLK} = \frac{\sin\frac12\stackrel{\frown}{BL}}{\sin\frac12\stackrel{\frown}{CK}} \\[6pt] \triangle CNP: &\quad\frac{|NC|}{|NP|}= \frac{\sin\angle CPL}{\sin\angle ACB} = \frac{\sin\frac12(\stackrel{\frown}{CL}+\stackrel{\frown}{AK})}{\sin\frac12\stackrel{\frown}{AB}} \end{align}$$

This gives the first two ratios in the six-ratio product shown in the question. The remaining ratios are handled similarly.

Blue
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