Let D be the reflection of B about line AC. Then the triangles
ACB and ACD are congruent with reverse orientation. Therefore,
D - A (B - A)'
------- = ----------
C - A (C - A)'
Hence,
C - A
D = A + ---------- (B - A)'
(C - A)'
Note: If triangles ABC and DEF are similar (with the same orientation), then
B - A E - D
------- = ------- or
C - A F - D
C - B F - E
------- = ------- or
A - B D - E
A - C D - F
------- = -------
B - C E - F
To see the first we must convert to polar form
B - A = c ei(theta) and
C - A = b ei(phi)
B - A c c
------- = --- ei(theta-phi) = --- ei(A)
C - A b b
Likewise,
E - D f
------- = --- ei(D)
C - A e
For similar triangles we have A = D and c/b = f/e.
For similar triangles with reverse orientation, we must reflect one of the triangles about the real axis. This is accomplished using complex conjugates giving triangle D'E'F' with
B - A E' - D' (E - D)'
------- = --------- = ----------
C - A F' - D' (F - D)'
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