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Consider $I$ being the incentre of $\triangle ABC$. $IF \perp AB$ ($F \in AB$). $AI$ extended intersects the circumcircle of $\triangle ABC$ at $D$ ($D \not\equiv A$) and $AD \cap BC = \{K\}$. $DF$ intersects the circumcircle of $\triangle ABC$ and $\triangle BKD$ respectively at $Q$ and $P$. Prove that $IP \perp CQ$.

I have proven that $KP \parallel CQ$.

Let $BC \cap QD = \{E\}$. We have that $ED \cdot EP = EB \cdot EK$ and $ED \cdot EQ = EB \cdot EC$.

$$\implies \dfrac{EP}{EQ} = \dfrac{ED \cdot EP}{ED \cdot EQ} = \dfrac{EB \cdot EK}{EB \cdot EC} = \dfrac{EK}{EC}$$

Using the intercept theorem for $\triangle ECQ$ and $\dfrac{EP}{EQ} = \dfrac{EK}{EC}$, we have that $KP \parallel CQ$.

I have tried to prove that $KP \perp PI$ by proving $\widehat{KHB} = \widehat{PIF}$ where $KP \cap AB = \{H\}$. But it didn't work.

I would be grateful if you could solve the problem.

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Let's prove that $IP \perp PK$, since indeed $PK || CQ$. (because $\angle DPK= \angle DBK =$ half the measure of arc CD $= \angle DQC$)

Consider an inversion centered at D with radius DB. I will denote by $T'$ the image of point $T$ under that inversion.

$B'=B$, $C'=C$, $I'=I$ (the latter is due to so-called "trillium lemma", that is, $DI=DB=DC$)

The line BC after the inversion becomes the circumscribed circle of $ABC$. So, $K' = A$.

The circle $BPKD$ becomes a line passing through $B'=B$ and $K' = A$ So, $P' = F$.

Now, we know that $F$ lies on the circle with diameter $AI$. The image of that circle under our inversion is the circle with diameter $KI$. We're done, since $F' = P$.

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