# Symmedian and Antiparallel

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A Mathematical Droodle

2 September 2015, Created with GeoGebra

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Copyright © 1996-2018 Alexander BogomolnyThe applet illustrates the following fact:

A symmedian drawn from a vertex of a triangle divides the antiparallels to the opposite side in half.

Indeed, let CS be the symmedian from vertex C and UV an antiparallel to the side AB. By the definition of the antiparallel, the triangles ABC and VUC are similar. Assume that this similarity maps CM to CT. Then the segments UT and VT are equal as are the angles ACM and VCT. It follows that CT is a part of the reflection of CM in the angle bisector of C. But this is exactly the definition of the symmedian. It follows that CS (CT extended) is the symmedian from C.

This result is often presented in a different form: the locus of the midpoints of the antiparallels to a side of a triangle is the summedian through the opposite vertex.

### Corollary

_{B}is antiparellel to both AB and BC. Therefore, it is halved by either of them. In other words, the midpoint of BH

_{b}lies on the symmedians through A and C. Since the symmedian point is the point of concurrency of the three symmedians, it is also the point where any two of them meet.

### 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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Copyright © 1996-2018 Alexander Bogomolny63437834 |