Chord-Chord Power Calculator
Intersecting Chords Theorem: a₁ × b₁ = a₂ × b₂
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Intersecting Chords Theorem
When two chords intersect inside a circle, the products of their segments are equal:
a₁ × b₁ = a₂ × b₂
Intersecting Chords
Understanding the Intersecting Chords Theorem
When two chords of a circle intersect at a point inside the circle, the products of their segments are equal. This elegant relationship is known as the Intersecting Chords Theorem or the Chord-Chord Power Theorem.
The Theorem
If two chords AB and CD intersect at point P inside a circle, then:
AP × PB = CP × PD
Or equivalently: a₁ × b₁ = a₂ × b₂
Why It Works
The proof uses similar triangles. Triangles APC and DPB are similar because:
- ∠APC = ∠DPB (vertical angles)
- ∠CAP = ∠BDP (inscribed angles on the same arc)
- From similarity: AP/DP = CP/BP
- Cross-multiplying: AP × BP = CP × DP
Power of a Point
The common product a₁ × b₁ = a₂ × b₂ is called the "power of the point P" with respect to the circle. This value is constant for any chord through P.
For a point at distance d from center of a circle with radius r:
Power = r² - d² (if inside) or d² - r² (if outside)
Frequently Asked Questions
Does this work for any two chords?
Yes, as long as both chords pass through the same point inside the circle. The theorem holds regardless of the chords' lengths or orientations.
What if the chords don't intersect inside the circle?
If the point is outside the circle, use the Secant-Tangent Power theorem instead: (whole₁)(external₁) = (whole₂)(external₂).
What is the power of a point?
The power of point P is the constant product obtained when any chord through P is divided into segments. It equals |r² - d²| where d is the distance from P to the center.
Can I use this to find a circle's radius?
Yes! If you know the chord segments and the distance from the intersection point to the center, you can use: r² = d² + (power of point).
Is this related to similar triangles?
Yes! The proof relies on showing that two triangles formed by the chords are similar, leading to the proportional relationship that gives us the theorem.