Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Tuesday, April 21, 2020

### Geometry Problem 1471: Equilateral Triangle, Inside/Outside Point, Incenters, Tangency Points, Concurrent Lines

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https://photos.app.goo.gl/4DmeeDEMfZsZ4ekS9

ReplyDeleteLet AB=BC=AC= a

Calculate PA^2-PB^2= a(P1A-P1B)

PB^2-PC^2= a(P2B-P2C)

PC^2-PA^2= a(P3C-P3A)

Add above expressions side by side and simplify ,we will have

(P1A-P1B)+ (P2B-P2C)+ (P3C-P3A)= 0…. (1)

Conversely , any 3 points P1, P2 and P3 satisfied expression (1) then the perpendiculars to the 3 sides of the equilar. Triangle will be concurred .

Let PA= u; PB=v; PC=w

We have T3C-T3A= w-u

Similarly (T1A-T1B)=u-v

And (T2B-T2C)=v-w

Add these above expression and found that the positions of T1, T2, T3 satisfy (1)

So the perpendiculars from these points to the sides will concur at a point D