Power of a Point Theorem

Step 1: Set up similar triangles. Let the two lines through $P$ meet the circle at $A,B$ and $C,D$. Since $A,B,C,D$ lie on a common circle, the inscribed angles $\angle PAC$ and $\angle PDB$ subtend the same arc $BC$ (or its supplement, depending on configuration), so $\angle PAC = \angle PDB$.

Step 2: Match a second pair of angles. The angle at $P$ in triangle $PAC$ and the angle at $P$ in triangle $PDB$ are either equal (if $P$ is outside the circle, the two lines share vertex $P$ so the angle is literally the same angle) or vertical angles (if $P$ is inside the circle). Either way $\angle APC = \angle DPB$.

Step 3: Conclude similarity. Two triangles $PAC$ and $PDB$ with two pairs of equal angles are similar by AA: $\triangle PAC \sim \triangle PDB$.

Step 4: Extract the ratio. Similar triangles give proportional corresponding sides: $\frac{PA}{PD} = \frac{PC}{PB}$, which rearranges to $PA \cdot PB = PC \cdot PD$ (using unsigned lengths first, for $P$ outside the circle).

Step 5: Compute the common value via a diameter. Choose the specific line through $P$ and the center $O$, meeting the circle at the two points on the diameter, at signed distances $OP - r$ and $OP + r$ from $P$ (taking $P$ outside so $OP > r$). Their product is $(OP-r)(OP+r) = OP^2 - r^2$, which by Step 4 must equal $PA \cdot PB$ for every line through $P$.

**Step 6: Handle $P$ inside the circle by signed lengths.** When $P$ lies inside the circle, $A,P,B$ are in that order on the chord, so $\overrightarrow{PA}$ and $\overrightarrow{PB}$ point in opposite directions and their signed product $\overrightarrow{PA}\cdot\overrightarrow{PB} = -PA\cdot PB$ is negative; repeating Steps 1–5 with the diameter through $P$ gives $-PA\cdot PB = OP^2-r^2$ (now negative since $OP<r$), so the identity $\overrightarrow{PA}\cdot\overrightarrow{PB}=OP^2-r^2$ holds uniformly with signs, completing the proof.