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 1115: Right Triangle, Angle Trisection, Equilateral

Labels:
angle,
equilateral,
right triangle,
trisection

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(Actually copy from my last solution)

ReplyDeleteSince AC₁ = AB₂.

So AA₁ is the perpendicular bisector of C₁B₂,

thus AC₁A₁B₂ is a kite, therefore A₁C₁=A₁B₂.

Similarly, C₁A₁=C₁B₂.

Hence, ΔB₂A₁C₁ is equilateral.

Another interesting fact:

ReplyDeleteB₁ is actually the center of equilateral ΔB₂A₁C₁

Reference my proof in the previous problem, A1B2C1 is an isoceles triangle with included angle = 60

ReplyDeleteHence this triangle has to be equilateral

Sumith Peiris

Moratuwa

Sri Lanka