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

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

Click the figure below to enlarge the figure of problem 1012.

## Monday, May 5, 2014

### Geometry Problem 1012: Equilateral Triangle, Incenters, Inscribed Circles, Perpendicular, Concurrent Lines

Labels:
circle,
concurrent,
equilateral,
incenter,
inscribed,
perpendicular,
triangle

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To prove that A1A2, B1B2, C1C2 concurrent at a point E we need:

ReplyDelete(C2B)^2 - (C2A)^2 + (B2A)^2 -(B2C)^2 + (A2C)^2 - (A2B)^2 = 0

Which is equivalent to C2B + B2A + A2C = C2A + B2C + A2B <=> C2B + B2A + A2C = 1.5a

We set AB = AC = BC = a

AD = x, BD = y; CD = z;

Since C2B = 0.5(y+a-x); B2A = 0.5(a+x-z); A2C = 0.5(a+z-y)

=> C2B + B2A + A2C = 1.5a

So: A1A2, B1B2, C1C2 concurrent at a point E