# M2_8_8_5_4

Author:
angela

## 8_5_4

Existence of the incenter. Prove that the interior angle bisectors of a triangle are concurrent.
Start with triangle ABC and angles A = α, B = β, C = γ and construct the angle bisectors are constructed which divides each angle in 2 equal parts. Now we have A = α/2 +α/2, B = β /2 + β/2, and C = γ/2 +γ/2. We must prove that the three angle bisectors of α, β, and γ are concurrent. If perpendicular lines are dropped from the center of the incircle to the three sides of ΔABC and label these points F, E, and D, where D is the point on BC, F is the point on AB, and E is the point on AC. We have now divided ΔABC into six smaller triangles. We will now prove that the six smaller triangles are in fact 3 pairs of congruent triangles. Look at ΔBID where I is the incenter. Right away we can see this triangle has the angle β/2 and a right angle. Hence, the third angle is 90 - β/2. Similarly, ΔBIF has angles β/2, 90 - β/2, and 90. Finally, we can see that ΔBID and ΔBIF share a hypotenuse BI. Therefore, by the Angle Side Angle Theorem the two triangles are congruent. Now look at ΔAIF. It has angles α/2, 90, and 90 - α/2 and by similar reasoning ΔAIE has these same angles. Like ΔBID and ΔBIF, these two triangles share a hypotenuse AI. Therefore, by the Angle Side Angle Theorem ΔAIF and ΔAIE are congruent. Finally, by a similar proof ΔCID and ΔCIE are congruent as well. In conclusion, because the angle bisectors of ΔABC form 3 pairs of congruent triangles and each hypotenuse of these congruent triangles ends at the point I we can conclude that the interior angle bisectors of ΔABC are concurrent at point I (the incenter). To explain using Ceva's Theorem: For angle bisectors we know that AF/FB = AC/BC, BD/DC = AB/AC, CE/EA = BC/AB.  Multiplying the three yields (1). Thus, Ceva’s Theorem applies and the three angle bisectors are concurrent.