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I believe the distance of a point in acute angle bisector is smaller than that of the obtuse angle bisector. I need to know if I know the angle between two lines is it possible to find the distance on the angle bisecenter image description heretor which will be equal to the distance on the line?

Let us say OA = OA` = OC all of same distance. But OC will not be perpendicular to OA due to the angle. I need to know what correlation can i come up the distance CD and angle so that i can find D to make it perpendicular to OA.

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What does "distance of a point" mean? Distance between a point and what? – Sammy Black Mar 17 '14 at 3:00
Hi Sammy, Sorry on that. I meant from the point of intersection if i find a point with distance 'd' i will find the point which will not be equal on the same distance if i calculate on the line. I need to know how the distance of the point varies from the intersecting point if i know the angle. – Raajesh Kotteeswaran Mar 17 '14 at 3:17
Wow! @RaajeshKotteeswaran , I'm afraid your "explanation" made things much really need to make an effort to make yourself clear, because so far you haven't succeeded. – DonAntonio Mar 17 '14 at 4:18
Hi DonAntonio, I agree since this needs more graphical representation to explain . I am trying to put one in place so that i can explain much better. – Raajesh Kotteeswaran Mar 17 '14 at 4:20
Way to go, @RaajeshKotteeswaran . – DonAntonio Mar 17 '14 at 4:31
up vote 1 down vote accepted

If $\phi / 2$ is the little angle at $O$, then by one definition of cosine in a right triangle we have $$\cos\left( \frac{\phi}{2} \right) = \frac{|OA|}{|OD|}$$ so $$\cos\left( \frac{\phi}{2} \right) = \frac{|OC|}{|OD|} \quad\text{ or }\quad |OD|=\frac{|OC|}{\cos\left( \frac{\phi}{2} \right)}$$ This shows how the relative distances (from $O$) of $C$ and $D$ depend on the angle.

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Thanks @Nielsen. – Raajesh Kotteeswaran Mar 19 '14 at 9:11
@RaajeshKotteeswaran Note that $1/(\cos x)$ is sometimes called $\sec x$ (the secant function of $x$), so you can also write $|OD|=|OC|\sec(\phi / 2)$. – Jeppe Stig Nielsen Mar 19 '14 at 10:27

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