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Assignment 4:

"Prove the the 3 angle bisectors of the inner angles that a

triangle room concurrent."

by

Margo Gonterman

The angle bisector of an angle abc is a ray ad such edge ABD = angle DBC.

You are watching: Point of concurrency of the angle bisectors of a triangle

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Angle Bisector Construction

- mark a suggest E top top the segment BC.

- build a circle with facility B and also radius BE.

- note the intersection the the circle with ab as allude F.

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- build a circle with center E and also radius EF.

- construct a one with center F and radius EF.

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- mark an intersection the the 2 circles as allude D.

- draw a ray AD.

- advertisement is the edge bisector of edge ABC.

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Proof of angle Bisector Construction

- attach EF, DE, and also DF.

- DF=FE due to the fact that both space radii the the circle with facility F.

- EF=EF since both space radii the the one with center E.

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- therefore EF=FD=DE.

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

- BF=BE through construction

- hence Triangle BEF is an isosceles triangle

- because the triangle is isosceles, angle BFE = edge BEF.

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

- edge BFD = edge BED

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

- Specifically, angle FBD = angle DBE

or angle ABD = edge DBC

- thus BD is the edge bisector of angle ABC

Proof of Concurrency of edge Bisectors the a Triangle

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

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

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

- brand the intersection point out on side BC, AC, abdominal as D, E, F respectively.

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- Now think about triangle BFK and triangle BDK.

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

- angle FBK = angle DBK due to the fact that BK is the edge bisector.

- BK is common in both triangles.

- Therefore, triangle BDK is congruent come triangle BFK through AAS.

- special, FK = DK.

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- Now take into consideration triangle AFK and triangle AEK.

- angle AFK = angle AEK because they are both appropriate angles.

- angle FAK = angle EAK due to the fact that AK is the angle bisector.

- AK is common to both triangles.

- as such triangle AFK is congruent to triangle AEK by AAS.

- specifically, FK= KE.

- FK = KE = KD.

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- Now, think about triangle CDK and also triangle CEK.

- KE=KD

- KC is typical to both triangles.

- angle KDC = edge KEC since they space both best angles.

- then triangle CDK is congruent to triangle CEK through Hypotenuse-Side

Click here to check out a evidence of Hypotenuse-Side congruence

- therefore angle DCK = angle ECK

Thus the edge bisectors of the interior angles that the triangle space concurrent.

See more: What Distinguishes One Element Is Distinguished From Another By The Number Of

Incenter

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