Prove that $AI=PI$

Question

Given a triangle $ABC$ with incenter $I$, line $AI$ intersects the circumcircle of triangle $ABC$ at points $A$ and $S$. Let $J$ be the resulting point after reflecting $I$ w.r.t. line $BC$, and let $SJ$ meets the circumcircle of triangle $ABC$ at points $S$ and $P$. Prove that $AI=PI$.

Can anyone help me solve this? I tried solving this for a long time but I am stuck nowhere. I knew that $S$ is the center of the circumcircle of triangle $BIC$, $\bigtriangleup BIJ \sim \bigtriangleup SIC$, and $\bigtriangleup CIJ \sim \bigtriangleup SIB$, but I still couldn't manage to show that $\angle PAI = \angle API$ by angle chasing. Any help will be really appreciated ! Thanks!

Answer 1

Extend PO to meet the circle ABC at Q. Then, $\angle PAQ = 90^0$.

Extend PI to meet AQ at R. Then, PR is the diameter of the circle PAR. Note that AP is the common chord of circles ABC and PAR.

Let T be the midpoint of AP. Then $\angle OTA = 90^0$.

It remains to argue why OIT is the line of centers.

Answer 2

We need the circumcenter $O$, it will join the darts. Here is a picture suggesting the proof, try to find it first using own devices.

Let us show first that the following triangles are similar: $$ \begin{aligned} \Delta & AIO\ ,&&(1)\\ \Delta & IJS\ . \end{aligned} $$ The blue angle is easy, $ \widehat{JIS}= \widehat{ISO}= \widehat{OAS}= \widehat{OAI} $, because $IJ\|OS$ and $\Delta AOS$ isosceles. We need one proportionality, so let us show $(!)$: $$ \frac{AI}{IJ} \overset{(!)}= \frac{AO}{IS}\ . \qquad(*) $$ Some elements can be quickly rewritten, using usual notations, $IJ=2r$, $AO=R$. So we need $AI\cdot IS$, the power of $I$ with respect to the circumcenter, to be $2rR$. (This relation is the way we prove the Euler formula. Assuming it, we are done with $(*)$, but since all ingredients are in the picture, let us proceed explicitly.) Let $F$ be the projection of $I$ on $AB$. It is easy to show that the triangles $\Delta AIF$ and $\Delta TBS$ are similar. (Right angles, and $\widehat{FAI}=\widehat{BAS}=\widehat{BTS}$.) We write the proportionality of the sides, $$ \frac{AI}{TS} = \frac{FI}{BS}\ , \qquad\text{ equivalently: } \frac{AI}{2R} = \frac{r}{IS}\ . $$ This shows the needed proportionality $(*)$.

It follows the equality of the green angles in the two similar triangles $(1)$, and we finish in the line $$ 2\widehat{PSA} = \overset\frown{PA} = \widehat{POA} = \widehat{POI} + \widehat{IOA} \ . $$ Explicitly: This implies that the two angles in the last sum are equal, so $OI$ is the angle bisector of $\widehat{POA}$ in the isosceles triangle $\Delta POA$, so $OI$ is the perpendicular bisector of the side $PA$, and $I$ being on this side we conclude $$ IP=IA\ . $$ $\square$

Bonus: The points $P,I,O,S$ are on a circle. (Because of the green angles in $O,S$.) Possibly there is a proof using this "coincidence", and involving only angles.