Prove that $AP \cdot AQ+CP \cdot CQ=BP\cdot BQ$

Question

Let $G$ be the centroid of $\triangle ABC$. A line $M$ through $G$ intersects the circumcircle of $\triangle ABC$ at $P$ and $Q$, where $A$ and $C$ lie on same side of $M$. Prove that $AP \cdot AQ +CP \cdot CQ=BP\cdot BQ$.

Answer 1

If we can prove that $[BPQ] = [PAQ] + [PCQ]$ we are done, because the right-hand side $$= (1/2)\cdot (\sin X)\cdot (AP\cdot PB + CP\cdot CQ),$$

and the left-hand side $$= (1/2)\cdot \sin(180 - X)\cdot (BP\cdot PQ).$$

Construction:

1. Let $BD$ be the median through $B$.

2. Draw $PD$ and $QD$.

Proof:

Now, $[PDQ]:[BPQ] = 1:2$, since $BG/GD = 2:1.\quad\quad\quad\quad\quad(1)$

Its trivial to prove that $[PDQ]:([PAQ] + [PCQ]) = 1:2.\quad\quad\quad(2)$

From Equations.(1) & (2), we get our result.