Steiner's Ratio Theorem
Let $D\;$ be a point on the sideline $BC\;$ of $\Delta ABC,\;$ and the reflection of the line $AD\;$ in the internal angle bisector of the angle $A\;$ intersect the line $BC\;$ at a point $E.\;$
Then
$\displaystyle\frac{BD}{CD}\cdot\frac{BE}{CE}=\frac{AB^2}{AC^2}.$
Lines $AD\;$ and $AE\;$ are said to be isogonal (or isogonal conjugate).
Proof
By the Law of Sines,
$\displaystyle\frac{BD}{AB}=\frac{\sin\angle BAD}{\sin\angle ADB}$ and $\displaystyle\frac{CD}{AC}=\frac{\sin\angle CAD}{\sin\angle ADC}=\frac{\sin\angle CAD}{\sin\angle ADB}.$
Dividing the two term-wise gives $\displaystyle\frac{BD}{CD}=\frac{AB}{AC}\frac{\sin\angle BAD}{\sin\angle CAD}.$
Similarly, $\displaystyle\frac{BE}{CE}=\frac{AB}{AC}\frac{\sin\angle BAE}{\sin\angle EAC}.$
The product of the two reduces to the required $\displaystyle\frac{BD}{CD}\cdot\frac{BE}{CE}=\frac{AB^2}{AC^2}$ because, by the construction, $\angle BAD=\angle EAC\;$ and $\angle BAE =\angle CAD.\;$
Corollary
Point $X\;$ on the side $BC\;$ of $\Delta ABC\;$ has the property that $\displaystyle\frac{BX}{CX}=\frac{AB^2}{AC^2}\;$ iff $AX\;$ is the symmedian through vertex $A.$
References
- Sammy Luo and Cosmin Pohoata, Let's Talk About Symmedians!, Mathematical Reflections 4 (2013), 1-11
Symmedian
- All about Symmedians
- Symmedian and Antiparallel
- Symmedian and 2 Antiparallels
- Symmedian in a Right Triangle
- Nobbs' Points and Gergonne Line
- Three Tangents Theorem
- A Tangent in Concurrency
- Symmedian and the Tangents
- Ceva's Theorem
- Bride's Chair
- Star of David
- Concyclic Circumcenters: A Dynamic View
- Concyclic Circumcenters: A Sequel
- Steiner's Ratio Theorem
- Symmedian via Squares and a Circle
- Symmedian via Parallel Transversal and Two Circles
- Symmedian and the Simson
- Characterization of the Symmedian Point with Medians and Orthic Triangle
- A Special Triangle with a Line Through the Lemoine Point
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