Assignment 4:

"Prove that the three angle bisectors of the internal angles of a

triangle are concurrent."

by

Margo Gonterman

The angle bisector of an angle ABC is a ray AD such angle ABD = angle DBC.

Angle Bisector Construction

- Mark a point E on the segment BC.

- Construct a circle with center B and radius BE.

- Mark the intersection of the circle with AB as point F.

- Construct a circle with center E and radius EF.

- Construct a circle with center F and radius EF.

- Mark an intersection of the two circles as point D.

- AD is the angle bisector of angle ABC.

Proof of Angle Bisector Construction

- Connect EF, DE, and DF.

- DF=FE since both are radii of the circle with center F.

- EF=EF since both are radii of the circle with center E.

- Therefore EF=FD=DE.

- Triangle DEF is an equilateral triangle and angle DEF = angle EFD = angle FDE = 60 degrees.

- BF=BE by construction

- Thus Triangle BEF is an isosceles triangle

- Since the triangle is isosceles, angle BFE = angle BEF.

- angle BFE + angle EFD = angle BEF + angle DEF

- angle BFD = angle BED

- Then triangle BFD is congruent to triangle BED by SAS.

- Specifically, angle FBD = angle DBE

or angle ABD = angle DBC

- Thus BD is the angle bisector of angle ABC

Proof of Concurrency of Angle Bisectors of a Triangle

- Given a triangle ABC, construct the angle bisectors of angle BAC and angle ABC.

- Mark the intersection of of the two rays as the point K.

- From the point K, construct segments that are perpendicular to each of the three sides of the triangle.

- Label the intersection points on side BC, AC, AB as D, E, F respectively.

- Now consider triangle BFK and triangle BDK.

- angle BDK = angle BFK since they are both right angles.

- angle FBK = angle DBK since BK is the angle bisector.

- BK is common in both triangles.

- Therefore, triangle BDK is congruent to triangle BFK by AAS.

- Specifically, FK = DK.

- Now consider triangle AFK and triangle AEK.

- angle AFK = angle AEK since they are both right angles.

- angle FAK = angle EAK since AK is the angle bisector.

- AK is common to both triangles.

- Therefore triangle AFK is congruent to triangle AEK by AAS.

- Specifically, FK= KE.

- FK = KE = KD.

- Now, consider triangle CDK and triangle CEK.

- KE=KD

- KC is common to both triangles.

- angle KDC = angle KEC since they are both right angles.

- Then triangle CDK is congruent to triangle CEK by Hypotenuse-Side

- Therefore angle DCK = angle ECK

Thus the angle bisectors of the interior angles of the triangle are concurrent.

Incenter

The point of concurrency of the angle bisectors of a triangle is the incenter of the triangle.