A circle intersects an equilateral triangle ABC in
six points, D, E, F, G, H, J. In traversing the perimeter
of the triangle, these points occur in the order A,
D, E, B, F, G, C, H, J.
Prove that AD+BF +CH = AJ + BE + CG.
It's "nearly trivial" in analytic geometry, by which I mean no square roots are needed.
Use an equilateral triangle of side length 1. Then
A = (0,0) and B = (1,0), and C = (½, ½√3).
Parameterize the circle by its center K = (m, n) and radius, r.
Parameterize D, E, F, G, H and J in terms of the fractional distance counter-clockwise about each side of the triangle. Then you have six variables d, e, f, g, h and j which are between zero and one. And since each side is of length 1, the fraction is the same as the length of the segment from the "origin vertex".
For example
AD = d
AE = e thus BE = 1 - e
Thus AD+BF +CH = AJ + BE + CG if and only if d + f + h = (1 - e) + ( 1 - g ) + (1 - j ) if and only if d + e + f + g + h + j = 3.
Since D and E are points on the circle, d and e are both solutions to the quadratic equation which you get from squaring both sides of
| (B-A)s - (K-A) | = r
( (B_x-A_x)s + (K_x-A_x) )² + ( (B_y-A_y)s + (K_y-A_y) )² - r² = 0
(B_x-A_x)²s² - 2(B_x-A_x)(K_x-A_x)s + (K_x-A_x)² + (B_y-A_y)²s² - 2(B_y-A_y)(K_y-A_y)s + (K_y-A_y)² - r² = 0
[(B_x-A_x)²+ (B_y-A_y)²]s² - 2[ (B_x-A_x)(K_x-A_x)+(B_y-A_y)(K_y-A_y) ]s + [ (K_x-A_x)² + (K_y-A_y)² - r² ] = 0
But since AB = 1 this is just
s² - 2[ (B_x-A_x)(K_x-A_x)+(B_y-A_y)(K_y-A_y) ]s + [(K_x-A_x)² + (K_y-A_y)² - r² ] = 0
and since we need the sum of both roots, we don't need to take the square root and
d + e = 2[ (B_x-A_x)(K_x-A_x)+(B_y-A_y)(K_y-A_y) ]
and so
d + e + f + g + h + j
= 2[ (B_x-A_x)(K_x-A_x)+(B_y-A_y)(K_y-A_y) ] + 2[ (C_x-B_x)(K_x-B_x)+(C_y-B_y)(K_y-B_y) ] + 2[ (A_x-C_x)(K_x-C_x)+(A_y-C_y)(K_y-C_y) ]
= 2[ (1-0)(K_x-0)+(0-0)(K_y-0) ] + 2[ (½-1)(K_x-1)+(½√3-0)(K_y-0) ] + 2[ (0-½)(K_x-½)+(0-½√3)(K_y-½√3) ]
= 2[ K_x ] + 2 [ ½ - ½ K_x + ½√3 K_y ] + 2[ ¼ -½ K_x + ¾ - ½√3 K_y ]
= 2 [ ½ + ¼ + ¾ ]
= 3
Thus AD+BF +CH = AJ + BE + CG
Use an equilateral triangle of side length 1. Then
A = (0,0) and B = (1,0), and C = (½, ½√3).
Parameterize the circle by its center K = (m, n) and radius, r.
Parameterize D, E, F, G, H and J in terms of the fractional distance counter-clockwise about each side of the triangle. Then you have six variables d, e, f, g, h and j which are between zero and one. And since each side is of length 1, the fraction is the same as the length of the segment from the "origin vertex".
For example
AD = d
AE = e thus BE = 1 - e
Thus AD+BF +CH = AJ + BE + CG if and only if d + f + h = (1 - e) + ( 1 - g ) + (1 - j ) if and only if d + e + f + g + h + j = 3.
Since D and E are points on the circle, d and e are both solutions to the quadratic equation which you get from squaring both sides of
| (B-A)s - (K-A) | = r
( (B_x-A_x)s + (K_x-A_x) )² + ( (B_y-A_y)s + (K_y-A_y) )² - r² = 0
(B_x-A_x)²s² - 2(B_x-A_x)(K_x-A_x)s + (K_x-A_x)² + (B_y-A_y)²s² - 2(B_y-A_y)(K_y-A_y)s + (K_y-A_y)² - r² = 0
[(B_x-A_x)²+ (B_y-A_y)²]s² - 2[ (B_x-A_x)(K_x-A_x)+(B_y-A_y)(K_y-A_y) ]s + [ (K_x-A_x)² + (K_y-A_y)² - r² ] = 0
But since AB = 1 this is just
s² - 2[ (B_x-A_x)(K_x-A_x)+(B_y-A_y)(K_y-A_y) ]s + [(K_x-A_x)² + (K_y-A_y)² - r² ] = 0
and since we need the sum of both roots, we don't need to take the square root and
d + e = 2[ (B_x-A_x)(K_x-A_x)+(B_y-A_y)(K_y-A_y) ]
and so
d + e + f + g + h + j
= 2[ (B_x-A_x)(K_x-A_x)+(B_y-A_y)(K_y-A_y) ] + 2[ (C_x-B_x)(K_x-B_x)+(C_y-B_y)(K_y-B_y) ] + 2[ (A_x-C_x)(K_x-C_x)+(A_y-C_y)(K_y-C_y) ]
= 2[ (1-0)(K_x-0)+(0-0)(K_y-0) ] + 2[ (½-1)(K_x-1)+(½√3-0)(K_y-0) ] + 2[ (0-½)(K_x-½)+(0-½√3)(K_y-½√3) ]
= 2[ K_x ] + 2 [ ½ - ½ K_x + ½√3 K_y ] + 2[ ¼ -½ K_x + ¾ - ½√3 K_y ]
= 2 [ ½ + ¼ + ¾ ]
= 3
Thus AD+BF +CH = AJ + BE + CG
Oh, circles are so mind boggling. Think of them as an infinite number of straight lines. 
So an infinite number of intersections.
So an infinite number of intersections.
Oh, I read your question wrong. It doesn't relate. Anyway, remove intersections and put infinite tangents, even though it still wouldnt' relate to your question.