# Proof of the Power of a Point Theorem

I had to dig for a bit to find a complete proof for each part of the Power of a Point Theorem, so I thought it would be useful to compile my own proof.

The Power of a Point Theorem states: Given a point P and a circle C, any line through P that intersect C will create either one segment, S (on a tangent line), or two segments, S1 and S2 (on a secant line), such that S2 or S1S2 is constant.

To prove this, we must prove it for all possible lines through P intersecting C.

There are several scenarios wrapped up in this theorem. The trivial case is that P is on C, in which case S or S1 is 0 and so is the product. These are the remaining scenarios:

1. P is inside of C. The two lines for our proof are both secant lines.
2. P is outside of C, and the two lines are both tangent lines.
3. P is outside of C, and the two lines are both secant lines.
4. P is outside of C, and the two lines are one secant line and one tangent line.

The fourth scenario is the one I had the most difficulty finding a clear explanation for, and is the most difficult.

### Case 1: Secant Lines (internal)

Three of the proofs rely on demonstrating that a particular pair of triangles are similar. In this case, consider △PED and △PAB. ∠PED and ∠BAP both intercept arc BD, and are hence congruent. ∠ABP and ∠EDP both intercept arc AE, and are likewise congruent. Triangles with two congruent angles are similar, and so  △PED ~ △PAB. This leads to: $\triangle PED \sim \triangle PAB \implies \\ \frac{PE}{PA} = \frac{PD}{PB} \implies \\ PA \cdot PD = PB \cdot PE$

This is what we needed to prove.

### Case 2: Tangent Lines

The second scenario is the easiest, and is usually proven first and independent of the rest. Since PA and PB are tangents, we know that ∠CAP and ∠CBP are right angles. We also know that AC ≅ BC (since they’re both radii) and that BC ≅ BC. Since △ACP and △BCP are both right triangles and we know two sides, the third sides can be calculated using the Pythagorean Theorem, and so AP ≅ BP. If AP = BP, then AP2 = BP2, which is what we need to prove.

### Case 3: Secant Lines (external)

Again, we rely on similar triangles, this time △PEA and △PDB. We know that ∠DPB ≅ ∠EPA. ∠AEB and ∠ADB intercept arc AB, so they are congruent, and their linear pair angles (∠BDP and ∠AEP) are likewise congruent. Thus △PEA ~ △PDB, and: $\triangle PEA \sim \triangle PDB \implies \\ \frac{PE}{PA} = \frac{PD}{PB} \implies \\ PA \cdot PD = PB \cdot PE$

This is what we needed to prove.

### Case 4: Secant Line and Tangent Line

This time, our similar triangles will be △PAE and △PBA. ∠APE ≅ ∠BPA. We need to find one other pair of congruent angles, and we’re all set.

There is a theorem that the arc intercepted by a chord and a tangent is twice the size of the corresponding angle. I’m going to prove that, rather than simply stating it. Since ∠DAE is inscribed, it is half the measure of arc DE (I’ll leave that one stated, not proven). Arc AED is 180°, while ∠DAP is 90° (since AP is tangent to the diameter AD). Since DE is twice ∠DAE and AED is twice ∠DAP, it must be the case that AE is twice ∠EAP. ∠ABE is inscribed, so it is half of arc AE and is thus congruent to ∠EAP.

From there, we finish the same as above: $\triangle PAE \sim \triangle PBA \implies \\ \frac{PE}{PA} = \frac{PA}{PB} \implies \\ PA \cdot PA = PB \cdot PE$

This proves that, for all lines between P and C, the product of the two segments (secants) or the square of the single segment (tangents) is always constant.

Clio Corvid

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