Question

1. can some explain to me on the proof how
A'B'+B'C'=A'C' goes to this step
=[r^2/(DA*DB)]AB + [r^2/(DB*DC)]BC +
[r^2/(DA*DC)]AC then it goes to step
AB*CD+BC*AC=AC*BD

Answer 1

A'B'+B'C'=A'C' ok?
Now, by distance formula in inversion center D, we have
A'B'= r^2 AB/ (DA*DB)
B'C'= r^2 BC/(DB*DC)
A'C'=r^2 AC/(DA*DC)

Answer 2

Put all in, we have
\[\dfrac{r^2 AB}{DA*DB}+\dfrac{r^2BC}{DB*DC}=\dfrac{r^2AC}{DA*DC}
\]

Answer 3

as it suggests, just multiple both sides by \(\dfrac{DA*DB*DC}{r^2}\) , you get what you want.

Answer 4

Proof of A'B'= r^2 AB/(DA*DB):
We know that the inversion in circle center D radius r gives us
DA*DA'= r^2
and we have \(\triangle DAB\cong \triangle DB'A'\)
Hence \(\dfrac{DA}{DB'}=\dfrac{DB}{DA'}\)\(\rightarrow DB'=\dfrac{DA*DA'}
{DB}=\dfrac{r^2}{DB}\)

Same argument in similar triangles, we have
\(\dfrac{AB}{B'A'}=\dfrac{DA}{DB'}\)\(\rightarrow A'B'=\dfrac{AB*DB'}{DA}\)
replace \(DB'\) above
You have
\(A'B'=\dfrac{AB*r^2}{DA*DB}\)
proof end