# Symmedian and 2 Antiparallels

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A Mathematical Droodle

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Copyright © 1996-2018 Alexander BogomolnyThe applets illustrates the following statement:

In a triangle ABC the antiparallels to sides AB and AC that meet on the symmedian from C have equal lengths.

Let CS be the symmedian and MR and LN the two antiparallels in question that meet in point T on CS. Triangle RTN having equal base angles at R and N is isosceles. Therefore,

(1) | TN = TR. |

Draw the third antiparallel UV through T.

Similarly to the above, we have

TL = TV and

TM = TU.

However, as we know,

TU = TV.

Therefore

MR | = TM + TR |

= TL + TN | |

= LN. |

Note that we actually got a little more than claimed: the corresponding pieces of the equal antiparallels cut off by the symmedian are also equal.

By transitivity, the three antiparallels through the symmedian point all have equal lengths.

### 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 Bogomolny65651660