# Prove that $IP \perp CQ$. 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.

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 $$F' = P$$.