tangent line to circle Theorem

logistic_guy

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Prove the theorem by contradiction.

Theorem_1.png

Hint: Assume the line m\displaystyle m is not perpendicular to the the segment QP\displaystyle \overline{QP}. Then assume the line m\displaystyle m is not tangent to the circle.

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Are you going to show some work, or is this just going to be another list of problems to waste our time on?

-Dan
 
Professor Dan is in here😍it's an honor. But I don't think that you are serious in helping me.

Are you going to show some work,
Of course. I know that you are here not to help me but I will give you some work because it is very rare to see a reply from you since the exposure of my identity.

Hint: Assume the line m\displaystyle m is not perpendicular to the the segment QP\displaystyle \overline{QP}.
Here is my attempt. Okay, I assumed that it is not perpendicular, so what is the big deal of this assumption? I also provided a diagram of my assumption to show you how serious I am.

tangent.png
 
Non-tangent means secant, which means a proper triangle, which means no right angles.

The rest are technical details.
 
Welcome professor Stefan.

Non-tangent means secant, which means a proper triangle, which means no right angles.
The discussion is now about the non-perpendicular assumption. When I prove it is perpendicular by contradiction, then I use this fact and the second assumption to prove it is tangent (this part should involve the secant).

I used the first assumption which means I assumed that the line m\displaystyle m is not perpendicular to the segment QP\displaystyle \overline{QP}. I don't see any information that contradict this fact. What about you? Do you see?
 
@topsquark
You could not help me. If you are not good in Geometry, why not just stick with RELATIVITY and DEs?

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If we assume that the segment QP\displaystyle \overline{QP} is not perpendicular to the line m\displaystyle m, then there must be a point on m\displaystyle m say k\displaystyle k such that the segment Qk\displaystyle \overline{Qk} is perpedicular to m\displaystyle m.

tangent_2.png
In a right triangle the segment (QP)\displaystyle \left(\overline{QP}\right) that is opposite to the 90\displaystyle 90^{\circ} angle must be the longest leg. It is not the case here. If the segment QP\displaystyle \overline{QP} is longer than the segment Qk\displaystyle \overline{Qk}, the segment Qk\displaystyle \overline{Qk} must lie inside the circle which contradicts our assumption. Then, the segment QP\displaystyle \overline{QP} is perpedicular to the line m\displaystyle m.

Part one has been proved, but we have not yet proved that the line m\displaystyle m is tangent to the circle. That would be done in the next Episode.

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