Can you guide me to prove the following theorem? Thanks in advance!
Secant-Tangent Intersection Theorem
What it says: When a secant and tangent intersect at the point of tangency, the angles created at the point of intersection are half the measurement of the arcs they intersect.

What it means: Imagine a line crosses through a circle (secant) and another line runs alongside the circle (tangent). These lines meet at a point on the circle. The two angles at this point are half the measure of the arcs they create on the circle.

Question

Can you guide me to prove the following theorem? Thanks in advance!
Secant-Tangent Intersection Theorem
What it says: When a secant and tangent intersect at the point of tangency, the angles created at the point of intersection are half the measurement of the arcs they intersect.

What it means: Imagine a line crosses through a circle (secant) and another line runs alongside the circle (tangent). These lines meet at a point on the circle. The two angles at this point are half the measure of the arcs they create on the circle.

anonymous · Answer

What it looks like: http://i.imgur.com/IgfcqpT.png

ganeshie8 · Answer

Join $SN$,
join $NL$

ganeshie8 · Answer

Look at triangle, $\triangle SNL$

ganeshie8 · Answer

can u find the angle, $\angle SNL$  in terms of the central angle, $\angle NSL$ ?

ganeshie8 · Answer

u may use "triangle sum property" and "isosceles triangle"

ganeshie8 · Answer

Since $\triangle SNL$ is an isosceles triangle, its base angles are cnongruent :

$\large \angle SNL \cong \angle SLN$

anonymous · Answer

Okay. Since all of the angles in a triangle add up to 180, and since angle SLN has to be congruent to angle SNL because it is isosceles, then it would be (180-(angle NSL))/2.

ganeshie8 · Answer

By triangle sum property we have :

$\large \angle SNL + \angle SLN + \angle NSL = 180$

ganeshie8 · Answer

Yes ! you have it,

$\large 2\angle SNL = 180 - \angle NSL$

$\large \angle SNL = \frac{180 - \angle NSL}{2}$

ganeshie8 · Answer

fine, so far eh ?

anonymous · Answer

yeah, i think i have an idea of where you're going to take this

anonymous · Answer

but im not completely sure, so lets go on

ganeshie8 · Answer

good :)

ganeshie8 · Answer

Next, we knw that tangent and radius intersect at 90 degrees, 
so we have $\large \angle SNE   \cong  90$

ganeshie8 · Answer

however $\large  \angle SNE \cong \angle SNL + \angle LNE$

ganeshie8 · Answer

so, 

$\large  90 \cong \frac{180 - \angle NSL}{2} + \angle LNE $

ganeshie8 · Answer

you can solve $\angle LNE$

anonymous · Answer

Okay.
$$180=(180-\angle NSL)+2\angle LNE \ 0=-\angle NSL + 2\angle LNE \ \angle NSL = 2\angle LNE \ \angle LNE = \frac{ 1 }{ 2 }\angle NSL$$

anonymous · Answer

wow, thank you

ganeshie8 · Answer

looks great ! u wlc :)