Prove that the three angle bisectors of the internal angles of a triangle are concurrent.
Here we have ΔABC with angles A = α, B = β, C = γ. The angle bisectors are constructed. Therefore, each angle is divided in two. Now 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 AB, F is the point on AC, and E is the point on BC. 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 ΔBGD where G 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, ΔBGE has angles β/2, 90 - β/2, and 90. Finally, we can see that ΔBGD and ΔBGE share a hypotenuse BG. Therefore, by the Angle Side Angle Theorem the two triangles are congruent.
Now look at ΔAGD. It has angles α/2, 90, and 90 - α/2 and by similar reasoning ΔAGF has these same angles. Like ΔBGE and ΔBGD, these two triangles share a hypotenuse AG. Therefore, by the Angle Side Angle Theorem ΔAGD and ΔAGF are congruent.
Finally, by a similar proof ΔCGF and ΔCGE 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 G we can conclude that the interior angle bisectors of ΔABC are concurrent.
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