Definition: A tangent line to a circle is a line that touches the circle at only one point.

Theorem: If some straight line touches a circle, i.e. is tangent to the circle, and some other straight line the circle to the point of contact, then the straight line, so joined, Proof

Let DE be tangent to the circle ABC at C. Let F be the center of ABC. Let FC be the radius in question. Suppose FC were not perpendicular to DE. Instead, suppose FG were drawn perpendicular to where FG intersects circle ABC. Then FGC is a right triangle, and by the Pythagorean Theorem, we have that FG 2 + GC 2 = FC 2 . Hence,

FG 2 < FC 2 as GC denotes a positive length.

FG 2 – FC 2 < 0

( FG – FC ) ( FG + FC ) < 0

( FG – FC ) < 0.

So FG < FC .

But FC = FB as both are radii of the same circle. So it follows from the above two relations that But this is impossible, since G lies outside of So FG cannot be perpendicular to DE. In the same way, it can be proven that no straight line through Therefore, FC t DE.

to a circle is a line that touches the circle at only one point.

touches a circle, i.e. is tangent to the circle, and some other straight line is joined from the center of , then the straight line, so joined, will be perpendicular to the tangent line.

were drawn perpendicular to DE. Call B be the point

is a right triangle, and by the Pythagorean Theorem, we have that

s a positive length.

as both are radii of the same circle.

that FG < FB.

of the circle ABC which is centered at F.

In the same way, it can be proven that no straight line through F except for FC can be perpendicular to DE

is joined from the center of

DE.

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INTERIOR ANGLE SUM THEOREM: The sum of the measures of the interior angles of any polygon with 𝑛 sides is

(𝑛 − 2) × 180° . In order to prove this theorem, we first prove something quite similar, but different. Namely, that any polygon with 𝑛 vertices (a polygon with 𝑛 vertices of course has 𝑛 sides) can be divided into a collection of (𝑛 − 2) triangles which, when put together, fill the entire polygon. All of the triangles in this collection have as their vertices points that are vertices of the original polygon as well. (See the figure for reference.) Let's assume that we know that such a triangulation always exists. Then, since each interior angle of the polygon is the sum of angles of the triangles, the total sum of all interior angles of the polygon is the same as the total sum of all interior angles of the triangles. But we know that the sum of angles of a triangle is 180o. So the sum of all angles of the polygon is equal to the number of triangles times the sum of the angles in each, which is equal to (𝑛 − 2) × 180°. Now, we prove that each polygon can be triangulated by induction. First, we show that each non-triangle polygon with vertices A, B, C,..., X has an "interior diagonal". That's a line segment from one vertex to another which never crosses the boundary of the polygon. (In the picture, BF is an interior diagonal, while BD would not be.) Take any convex angle in the polygon (A convex angle is one with measure of less than 1800. There must be one, since, for example, there's a rightmost point B in the polygon; since both of the vertices A and C that are adjacent to B are to its left, the angle :ABC is convex, (i.e. ;ABC < 180°). Then one of two things is true. Either AC is enclosed by the polygon, at which point AC is our diagonal, or it's not. Suppose AC crosses the boundary of the polygon. (This is actually the case in the figure.) Then there must be a vertex of the polygon inside the triangle ABC. In fact, if we sweep lines parallel to AC from B to AC, there will be a first–crossed vertex inside the triangle (it's F in our example). However, since it's the first–crossed vertex, that means that the triangle from B to the intersection of our line with AB and BC will be entirely contained inside the polygon. (In the picture, this is the triangle x1 x2 B.) So the line from B to this vertex is entirely inside the triangle, and thus inside the polygon: it's an internal diagonal.

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So any non-triangle polygon has an internal diagonal. Finally, we prove our triangulation theorem: Any polygon of 𝑛 vertices can be triangulated.

Base Case: 𝑛 = 3 is trivial.

Inductive Hypothesis: Any polygon with less than n vertices can be triangulated.

Induction Case: Given our 𝑛 − 𝑔𝑜𝑛, P, there exists an internal diagonal AB that splits the polygon into two polygons, Q and R. If Q has 𝑘 vertices, then R has (𝑛 – 𝑘 + 2) vertices, since we reuse A and B. But our inductive hypothesis

says that Q can be triangulated into (𝑘 − 2) triangles and R into (𝑛 − 𝑘) triangles. Since P is the disjoint union of

Q and R (except for boundaries), P will be the union of all the triangles of Q and R, which don't intersect. So there will be:

(𝑘 – 2) + (𝑛 – 𝑘) = (𝑛 − 2) triangles, all together,

and the induction holds. Therefore, any polygon P of 𝑛 vertices can be triangulated into (𝑛 − 2) triangles, and the formula works as desired for the sum of the interior angles.

3.20