Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Thursday, April 30, 2015

### Geometry Problem 1116: Right Triangle, Angle Trisection, 90 Degrees, Perpendicular Lines

Labels:
90,
perpendicular,
right triangle,
trisection

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http://s23.postimg.org/v3o05277v/pro_1116.png

ReplyDeletePer the result of problem 1114 we have

∠ (AB1C)= 120 degrees and B1B2 is an angle bisector of ∠ (AB1C)

And B2 is the incenter of triangle AB1C.

This problem will become problem 368 .

See link below for the solution.

http://gogeometry.blogspot.com/2009/10/problem-368-triangle-120-degrees-angle.html

Since B₂ is the in-center of ΔACB₁, thus we have

ReplyDeleteAC₃ / C₃B₁ = AC / B₁C = sin120° / sin(2A/3) = sin60° / sin(2A/3) = AB₃ / B₁B₃

therefore B₃C₃ is the angle bisector of ∠AB₃B₁.

Similarly, B₃A₃ is the angle bisector of ∠CB₃B₁.

As a result, ∠A₃B₃C₃ = 90°.

Kindly refer my proofs for Problems 1114 & 1115.

ReplyDeleteIn the given diagram all 4 acute angles around B1 have been shown to be 60. Hence A3 is the ex centre of Tr. AB1B2 and C3 of Tr. CB1B2. So C3B3 and A3B3 are external bisectors of the remaining angles of these Tr .s and the result follows.

Sumith Peiris

Moratuwa

Sri Lanka