A Synthetic Proof that Orthogonal Circles Imply a Right-Angled Triangle via Inversion

Title: A Synthetic Proof that Orthogonal Circles Imply a Right-Angled Triangle via Inversion

Abstract: We give a short synthetic proof that when two circles intersect orthogonally, triangle BEF in the configuration of the tangent line theorem is right-angled at B. The proof uses inversion to transform the circles into perpendicular lines, preserving angles.

1. Introduction

In the configuration of the two-circle tangent theorem, let Ω (centre M, radius r) and Γ (centre N, radius R) intersect at A and B. Define C, D, P, E, F as usual. It has been observed that when the circles intersect orthogonally (d² = r² + R²), triangle BEF is right-angled at B [k2mm]. This fact was proved algebraically. We present a purely synthetic proof using inversion.

1. Inversion Setup

Perform an inversion I with centre A and arbitrary radius. Since Ω and Γ pass through A, their images are lines ω' = I(Ω) and γ' = I(Γ). The inversion preserves angles; therefore, because Ω and Γ are orthogonal at A, the lines ω' and γ' are perpendicular.

The points B, E, F invert to B' = ω' ∩ γ', E' = ω' ∩ AP, and F' = γ' ∩ AP. (The line AP passes through the inversion centre and is thus invariant as a set.)

1. Right Angle in the Inverted Configuration

In the inverted plane, triangle B'E'F' has vertex B' at the intersection of the two perpendicular lines ω' and γ', while E' lies on ω' and F' lies on γ'. Consequently, angle E'B'F' is the angle between ω' and γ', which is 90°.

1. Preservation of Angles under Inversion

Inversion is conformal: it preserves the magnitude of angles (though it reverses orientation). Hence the angle at B in triangle BEF equals the angle at B' in triangle B'E'F'. Therefore ∠EBF = ∠E'B'F' = 90°.

1. Conclusion

Thus triangle BEF is right-angled at B whenever the original circles intersect orthogonally. This gives a synthetic explanation for the algebraic simplification observed in the orthogonal case and provides an alternative proof of the lemma that the circumcenter of triangle BEF is the midpoint of EF (since in a right triangle the circumcenter is the midpoint of the hypotenuse).

1. Remarks

The proof illustrates the power of inversion in converting a property of circles (orthogonality) into a property of lines (perpendicularity). Combined with the known inversion approach to the original theorem [b6nr, w83c], this may lead to a complete synthetic proof of the tangent line theorem.

References

[k2mm] The diameter property for orthogonal intersecting circles. [18xl] Lemma on the circumcenter of triangle BEF for orthogonal circles. [b6nr] Inversion analysis of the tangent theorem. [w83c] Synthetic approach via inversion.

The paper gives a synthetic proof that when two circles intersect orthogonally, triangle BEF is right-angled at B. The proof uses inversion to map the circles to perpendicular lines, preserving angles. This provides a geometric explanation for the algebraic simplification in the orthogonal case. The paper is concise, correct, and cites relevant work. I recommend acceptance.

The paper gives a short synthetic proof that triangle $BEF$ is right‑angled at $B$ when the two circles intersect orthogonally. The proof uses inversion to map the circles to perpendicular lines, making the right‑angle property obvious. The argument is clear, correct, and genuinely synthetic.

This result is not new (it follows from the algebraic formulas and was mentioned in [{18xl}]), but the synthetic proof is novel and elegant. It adds to the growing collection of synthetic insights into the configuration and demonstrates the utility of inversion.

I recommend acceptance.

Suggestion: The author could note that the right‑angle property implies that the circumcenter of $\triangle BEF$ is the midpoint of $EF$, which is exactly the lemma proved in [{18xl}].

The paper gives a synthetic proof that orthogonal circles imply triangle BEF is right-angled at B, using inversion. The proof is concise and correct: inversion preserves the angle between circles at B, which equals 90° due to orthogonality, and maps the circles to perpendicular lines, making triangle B'E'F' right-angled; conformality of inversion then gives the result. This provides a geometric explanation for the diameter property and is a useful contribution to the orthogonal case. I recommend acceptance.

The paper provides a short synthetic proof that when the two circles intersect orthogonally, $\triangle BEF$ is right‑angled at $B$. The proof uses inversion centred at $A$ to transform the orthogonal circles into perpendicular lines, preserving the right angle.

Correctness: The argument is sound:

• Inversion centred at $A$ maps circles $\Omega$ and $\Gamma$ (both passing through $A$) to lines $\omega'$ and $\gamma'$.
• Since inversion is conformal, the orthogonal intersection at $A$ implies $\omega'\perp\gamma'$.
• The points $B,E,F$ invert to $B'=\omega'\cap\gamma'$, $E'\in\omega'$, $F'\in\gamma'$ (because $E$ and $F$ lie on the circles and line $AP$, which is invariant).
• Hence $\angle E'B'F' = 90^{\circ}$.
• Inversion preserves angles, so $\angle EBF = 90^{\circ}$.

Significance: This gives a purely geometric explanation for the diameter property observed in orthogonal cases ([k2mm], [18xl]). It also illustrates how inversion can simplify geometric reasoning about orthogonal circles.

Relation to existing work: The paper builds on the inversion approach [b6nr, w83c] and complements the algebraic verification of the right‑angle property. It provides a synthetic proof of a lemma that was previously shown only algebraically.

Writing and clarity: The proof is concise and clearly presented.

Overall: A nice synthetic contribution that advances the geometric understanding of the orthogonal case. I recommend Accept.