Problem 1661

1. Define Points and Setup:
- Let $S$ be the circle, and $P$ be a point outside $S$.
- The tangent lines from $P$ meet $S$ at points $A$ and $B$.
- Let $M$ be the midpoint of $AB$.
- The perpendicular bisector of $AM$ meets $S$ at point $C$, which lies inside the triangle $ABP$.
- Let $AC$ intersect $PM$ at $G$.
- Let $PM$ meet $S$ at point $D$, which lies outside the triangle $ABP$.
- Given that $BD \parallel AC$, we need to show that $G$ is the centroid of triangle $ABP$.

2. Midpoints and Parallel Lines:
- Let $Q$ be the midpoint of $AM$.
- Let $T$ be the intersection of the perpendicular bisector of $AM$ with $AP$.
- Since $Q$ is the midpoint of $AM$ and $QT \parallel PM$, $T$ is the midpoint of $AP$.

3. Properties of Midpoints and Parallelograms:
- Since $C$ is the midpoint of $AG$, we have $AC = CG$.
- This implies $MC \parallel BG$.
- Given $AM = MB$, we have $BG = AD$.
- Therefore, $ADGB$ forms a parallelogram, and $AG = GB = BD = DA$.

4. Angle Relationships:
- Since $\angle ADP = \angle PDB = \angle BGD = \angle DGA = \alpha$ and $\angle BAG = \angle BAD = \angle ABG = \angle ABD = 90^\circ - \alpha$, we have $\angle BAP = \angle ABP = 2\alpha$.

5. Cyclic Quadrilaterals and Angle Chasing:
- Let $R$ be the foot of the perpendicular from $B$ to $AG$.
- Let $X$ be the intersection of $PM$ with $S$.
- Since $\angle ABX = \angle ADX = \alpha$, points $R$, $X$, and $B$ are collinear.
- The quadrilateral $AMXR$ is cyclic, so $\angle AMR = \angle AXR = \angle ADB = 2\alpha$.

6. Isosceles Triangle and Perpendicular Bisector:
- Let $T_1$ be the intersection of $MR$ and $AP$.
- Since $AMT_1$ is isosceles, $T_1$ lies on the perpendicular bisector of $AM$, so $T_1 = T$.
- Also, $\angle MRA = \angle MXA = 90^\circ - \alpha$, so $MR = AM$.

7. Cyclic Quadrilateral and Parallel Lines:
- The quadrilateral $QCRB$ is cyclic, so $\angle BQR = \angle BCR = 2\alpha$.
- This implies $QR \parallel AT$ (since $\angle BQR = \angle BAT = 2\alpha$).
- Since $QM = QA$ and $MR = RT$, we have $AT = AB$.

8. Intersection and Centroid:
- Let $T'$ be the point where $BG$ touches $AP$.
- By angle chasing, $\angle ABT' = \angle AT'B = 90^\circ - \alpha$, so $T' = T$.
- Therefore, $G$ is the intersection of $PM$ and $BT$, with $M$ and $T$ being midpoints of $AB$ and $AP$, respectively.
- Hence, $G$ is the centroid of triangle $ABP$.

$\blacksquare$