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Angle Bisector and Incenter of Triangle
Theorem: As shown below, is the bisector of BAC. Then
Proof:
1. From point C, draw a line such that the line is parallel to as shown above. In other words,
// . Thus, BAD = E and DAC = ACE .
2. Furthermore, is the bisector of BAC BAD = DAC
Hence, E = ACE . =
3. //
Finally, we have
Theorem : O is the incenter of ABC. Then O is equidistant to the three sides of ABC.
Proof: It can be shown by repeatedly using AAS congruence theorem.
Corollary: O is also the center of the inscribed circle of ABC. Then each side of the triangle is an tangent line of the circle. This circle is also known as “Incircle” .
Proof:
1. From the theorem, we know that O is equidistant to three sides. Thus, a circle exists with center O and the distance to one side as radius.
2. The three sides are perpendicular to its corresponding line segment of radius and the crossing point is on the other end of the radius. Thus, these three sides are tangent lines of the circle.
Corollary: Let r be the radius of the incircle of ABC. Then ABC = ½ ( + + ) r
Proof: It can be easily shown by using the diagram above of incircle.
Theorem: In the following diagram, O is the incenter of ABC.
Then = 1
Proof:
1. = , = , = 2, = 1 ( by using the equalities above ).
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