In triangle ABC, the bisectors of the exterior angles at B and C meet at D, and the bisectors of angle ABC and ACB meet at E. Prove points A, E, and D are collinear.

The statement is proved with the use of an auxiliary circle.

Expert Answers

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Please look at the picture. First, draw the line `AE `, which is known to be the bisector of the angle `A . ` Then denote `ABE = CBE = b , ` `ACE = BCE = c , ` so `BAE = pi / 2 - b - c .`

Then observe that `DBC = ( pi - 2b ) / 2 = pi / 2 - b = b' , ` `DCB = ( pi - 2c ) / 2 = pi / 2 - c = c' . ` Because of this, `DBE = b + b' = pi / 2 , ` `DCE = c + c' = pi / 2 .`

Now draw a circle which uses `DE ` as a diameter. Because `DBE ` and `DCE ` are right angles, `B ` and `C ` lie on this circle.

In a circle, a central angle is twice greater than the angle that stays on the same chord and has a vertex on the circle. This way, `BOD = 2 BCD = 2c' , ` so `BOE = pi - 2c' . ` And `BDE = 1 / 2 BOE = pi / 2 - c' = c .`

Now the sum of three angles of the quadrilateral `EDBA ` is `c + pi / 2 + b + pi / 2 - b - c = pi , ` which means `AED = 2pi - pi = pi , ` which is what we want.

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Last Updated by eNotes Editorial on February 19, 2021
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