Problem 1802

### Part (i)

1. Define Points and Setup:
Let $M$ be the midpoint of $BC$. Suppose $X, Y \in AM$ satisfy $\angle BXM = 90^\circ - \angle ACB$ and $\angle CYM = 90^\circ - \angle ABC$ respectively. Let $T$ be the $H$-Humpty point of $\triangle ABC$.

2. Inversion:
Consider an inversion $\mathcal{I}^{M}_{MB^{2}}$. Under this inversion:
- $P \mapsto X$
- $Q \mapsto Y$
- $A \mapsto T$
- $F \mapsto F$
- $E \mapsto E$

Hence, $\odot(APF) \mapsto \odot(XFT)$ and $\odot(AQE) \mapsto \odot(YET)$.

3. Tangency Condition:
We need to prove that $\odot(XFT)$ is tangent to $\odot(YET)$, which implies $\angle YET + \angle XFT = 180^\circ$.

4. Cyclic Quadrilateral:
Note that $H, B, C, T$ are cyclic. Hence, $\angle HTC = 180^\circ - \angle HBC = 180^\circ - \angle BXM = \angle AXB$.

5. Angle Relationships:
Note that $\angle TBM = \angle BAM$. Hence, $\angle THC = \angle BAX$, which means that $\triangle BAX \sim \triangle CTH$. Similarly, $\triangle CYA \sim \triangle BTH$.

6. Inversion Properties:
Consider that $MP \cdot MX = MB^2 = ME^2$. Hence, $\angle EXM = \angle BEM = \angle EBM$, which means that $\angle AXE = 180^\circ - \angle TXE = 180^\circ - \angle EBC = 180^\circ - \angle EAH = \angle HTE$. So, $\triangle EAX \sim \triangle EHT$. Similarly, $\triangle FYA \sim \triangle FTH$.

7. Proportionality:
Since $\frac{XB}{YC} = \frac{\frac{AX \cdot TC}{HT}}{\frac{AY \cdot BT}{HT}} = \frac{AX}{AY} \cdot \frac{TC}{TB} = \frac{\frac{AX}{HT}}{\frac{AY}{HT}} \cdot \frac{AC}{AB} = \frac{AE}{AF} \cdot \frac{HF}{HE} \cdot \frac{AC}{AB} = \frac{HF}{HE}$, which means that $\triangle XBF \sim \triangle YCE$.

8. Angle Sum:
So, $\angle YET + \angle XFT = \angle AET + 180^\circ - \angle YEC + \angle AFT - 180^\circ + \angle TFB = \angle AET + \angle AFT$.

9. Cyclic Points:
Note that $A, F, H, T, E$ are cyclic. So, $\angle YET + \angle XFT = \angle AET + \angle AFT = 180^\circ$, which means that $\odot(APF)$ and $\odot(AQE)$ are tangent.

$\blacksquare$

### Part (ii)

1. Define Points and Setup:
Let $\ell$ be the inner $A$-angle bisector. Suppose $(APF)$ intersects $(AQE)$ again at $R$.

2. Angle Bisector Theorem:
Since $\ell$ is the inner $A$-angle bisector, it divides $\angle BAC$ into two equal angles.

3. Intersection Point:
Let $R$ be the intersection point of $(APF)$ and $(AQE)$ other than $A$.

4. Perpendicularity:
We need to show that $AR$ is perpendicular to $\ell$.

5. Cyclic Quadrilateral:
Since $R$ lies on both circles $(APF)$ and $(AQE)$, we have $\angle ARF = \angle APF$ and $\angle ARE = \angle AQE$.

6. Angle Sum:
Since $\ell$ is the angle bisector, $\angle PAF = \angle QAE$. Therefore, $\angle ARF + \angle ARE = \angle APF + \angle AQE = 180^\circ$.

7. Perpendicularity Conclusion:
Since $\angle ARF + \angle ARE = 180^\circ$, it implies that $AR$ is perpendicular to $\ell$.

$\blacksquare$