\(\displaystyle \bigtriangleup\) ABC is similar to \(\displaystyle \bigtriangleup DEF\)

So \(\displaystyle \frac{AB}{AC} \equiv \frac{FD}{FE}\) is what you are used to

but this \(\displaystyle \leftrightarrow \frac{FE}{AC} \equiv \frac{FD}{AB}\) which you are not used to!

In many ways, I find the second way more understandable, as the lengths in

__one shape__ are the same fraction of the lengths in the

__other shape__, and this ratio represents just that fraction. E.g. the lengths in the second triangle might be 1/2 of the corresponding lengths in the first triangle. Therefore if you take the ratio of corresponding sides in the two shapes you should always get the same ratio, namely 1:2 or \(\displaystyle \frac{1}{2}\)

Next, this line:

The first part is simply comparing corresponding sides in the two similar triangles:

\(\displaystyle |P'Q'| : |PQ| \equiv |OP'|:|OQ| \hspace1ex\) (1)

They now simply multiply both parts of the ratio on the right hand-side by |OP| and therefore get an equivalent ratio -

\(\displaystyle (|P'Q'| : |PQ|) \hspace2ex \equiv \hspace2ex |OP'|:|OQ| \hspace2ex \equiv \hspace2ex |OP|.|OP'|:|OP|.|OQ| \hspace1ex \) (2)

(They could have multiplied by any positive number, but they chose |OP|)

Finally they use the fact that |OP|.|OP'| is \(\displaystyle k^2\) (since P' is the inverse of P in the circle) to rewrite the left-hand part of the final ratio in (2):

\(\displaystyle (|P'Q'| : |PQ| \hspace2ex \equiv \hspace2ex |OP'|:|OQ| \hspace2ex \equiv \hspace2ex |OP|.|OP'|:|OP|.|OQ|) \hspace2ex \equiv \hspace2ex |OP|.|OP'|:|OP|.|OQ|\hspace2ex \equiv k^2 : |OP|.|OQ| \hspace1ex\) (3)