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Let $C$ be a circle with the middle point $O$ and the radius $r$, we say that the points $P$ and $P'$ are inverse points with respect to $C$ if:

$1.$ $|OP|·|OP'|=r^2$

$2.$ $P$ and $P'$ are on the same line that starts in $O$

Assume that $P$ is located outside $C$ and draw the circle $C'$ with the diameter $\mid OP \mid$. $A$ is one of the intersection points between the circles $C$ and $C'$.Draw the normal line $N$ from $A$ towards $\mid OP \mid$ and let $P'$ be the point where $N$ intersects with $\mid OP \mid$ (see figure below). Prove that $P$ and $P'$ are inverse points with respect to $C$.

I drew $ΔOAP$ and $OAP'$ and proved that they are similar since they have 3 congruent angels and then I used that $\frac{|OA|}{|OP|}=\frac{|OP'|}{|OA|}$, where $|OA|$ is the radius in circle $C$ and then I got that $|OP|·|OP'|=r^2$, which is correct but that wasn't the proof that we were supposed to use so not really sure how to prove it in a different way.

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  • $\begingroup$ What have you attempted so far? $\endgroup$ Commented May 2, 2019 at 19:55
  • $\begingroup$ I drew the triangles OAP and OAP' and proved that they are Similiar since they have 3 congruent angels and then I used that |OA|/|OP|=|OP'|/|OA|, where |OA| is the radius in circle C and then I got that |OP|·|OP'|=r^2 which is correct but that wasnt the proof that we were supposed to use so not really sure how to prove it in a different way. $\endgroup$ Commented May 2, 2019 at 20:11
  • $\begingroup$ You should probably add that to your question. Comments are not permanent. $\endgroup$ Commented May 2, 2019 at 20:14
  • $\begingroup$ did, sorry havnt posted here before just read stuff :) $\endgroup$ Commented May 2, 2019 at 20:16

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Let B be the second point of intersection of $C$ and $C'$. Do an inversion in $C$. Points A and B are fixed under that inversion. The image of the circle $C'$ under that inversion is a line, since $C'$ passes through the center of inversion. That line contains points $A$ and $B$, so the image of $C'$ is the line $AB$ and the image of $P$ is $P'$

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  • $\begingroup$ How does that prove it? $\endgroup$ Commented May 3, 2019 at 6:56
  • $\begingroup$ @Jordan1333 I'm using here the following properties of inversion:$$ $$ 1) fixed points under inversion are exactly those lying on the circle of inversion (so, A an B are fixed) $$ $$ 2) the image of a circle passing through the center of inversion is a line not passing through the center (so, the image of the circle OAPB is AB; and so we know that the image of P lies on AB)$$ $$ 3) The image of P lies on OP. $$ $$ 1 and 3 are obvious, and 2 is one of the main facts about inversion. Actually, the proof of 2 uses similarity of triangles, I don't know why you want to avoid it $\endgroup$ Commented May 3, 2019 at 11:32

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