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Symmedian and 2 Antiparallels
What is this about?
A Mathematical Droodle
Explanation
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Copyright © 1996-2012 Alexander Bogomolny
The 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,
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. |
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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
|Activities|
|Contact|
|Front page|
|Contents|
|Geometry|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
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