Why are these two angles congruent?

[Noxide]

Why does angle LAB = angle ACB?

line AL is tangent to the circle at A
Triangle ABC is inscribed in the circle

http://img13.imageshack.us/img13/9553/tricircle.png (http://img13.imageshack.us/i/tricircle.png/)

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[berkeman]

Why does angle LAB = angle ACB?

line AL is tangent to the circle at A
Triangle ABC is inscribed in the circle

http://img13.imageshack.us/img13/9553/tricircle.png (http://img13.imageshack.us/i/tricircle.png/)

Uploaded with ImageShack.us (http://imageshack.us)

Is BC given as a diameter, or can it move up and down within the circle?

[Noxide]

Triangle ABC is any triangle.

My book says it's because both angles are measured from half of arc AB but I don't see how they can say that.

[LCKurtz]

I presume you see why angle ACB is half of arc AB. (The inscribed angle is half the corresponding central angle.) Consider a point A' just a very slight distance counterclockwise from A but really close to it on the arc, and consider the angle A'AB. This is an inscribed angle whose arc is almost arc AB so its angle is almost half of arc AB. As A' moves closer to A arc A'B moves closer to AB until when A' = A the subtended arc is AB and its angle is half of arc AB.

[Noxide]

I see that angle ACB is half of arc AB. But I don't see why we can also say that angle LAB is congruent to ACB because LAB is measured from half of arc AB (which is the reason my textbook gives for the congruence of ACB and LAB).

[zgozvrm]

Can you show angle LAC = angle LBA or triangle LAC congruent to triangle LBA?

[Noxide]

Can you show angle LAC = angle LBA...


I know that that statement is true, but only because my text states it, however my text does not explain why it is true. If you could explain why it is true or give me a theorem that would help, I would greatly appreciate it.

[LCKurtz]

Here's a new picture:
http://math.asu.edu/~kurtz/pix/forumcircleangles.jpg

You already know that angle c is on half arc AB = 1/2 angle o.

Notice that 2a + o = 180 degrees. Also, since the tangent line is perpendicular to the radius, b + a = 90 or 2b + 2a = 180.

So 2b + 2a = 2a + o or 2b = o so b = 1/2 o which is 1/2 arc AB so b = c.

[zgozvrm]

Here's a new picture:
http://math.asu.edu/~kurtz/pix/forumcircleangles.jpg

You already know that angle c is on half arc AB = 1/2 angle o.

Notice that 2a + o = 180 degrees. Also, since the tangent line is perpendicular to the radius, b + a = 90 or 2b + 2a = 180.

So 2b + 2a = 2a + o or 2b = o so b = 1/2 o which is 1/2 arc AB so b = c.

Nice!

[Noxide]

:approve:


Thank you!

[Noxide]

I presume you see why angle ACB is half of arc AB. (The inscribed angle is half the corresponding central angle.) Consider a point A' just a very slight distance counterclockwise from A but really close to it on the arc, and consider the angle A'AB. This is an inscribed angle whose arc is almost arc AB so its angle is almost half of arc AB. As A' moves closer to A arc A'B moves closer to AB until when A' = A the subtended arc is AB and its angle is half of arc AB.

Having seen your alternate explanation and proof, I am curious for which situations the above reasoning can be applied. Would this be considered logically sound in a formal proof (using limit arguments) or is this just an intuitive explanation? I fear that my inability to decipher this problem on my own was due to my lack of familiarity with the basic theorems of euclidean geometry. In fact, I went as far as trying to use the parallel lines associated with the given triangle's orthic triangle to alebraically determine the congruence of those two angles. I have to thank you again for this great solution.

[LCKurtz]

Having seen your alternate explanation and proof, I am curious for which situations the above reasoning can be applied. Would this be considered logically sound in a formal proof (using limit arguments) or is this just an intuitive explanation? I fear that my inability to decipher this problem on my own was due to my lack of familiarity with the basic theorems of euclidean geometry. In fact, I went as far as trying to use the parallel lines associated with the given triangle's orthic triangle to alebraically determine the congruence of those two angles. I have to thank you again for this great solution.

The point of my first post was to show you that the tangent case was just the limit of the inscribed angle case, so you would expect the same answer. I suppose you could make it mathematically rigorous but it would take more detail. But the fact that it is the limiting case of the inscribed angle suggests that the usual proof for the inscribed angle should lead the way to the proof in the tangent case. That is in fact how I came to the proof I gave.