# The Mirror Property of Altitudes via Pascal's Hexagram

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A Mathematical Droodle

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Copyright © 1996-2017 Alexander BogomolnyThe applet may suggest the following statement by Greg Markowsky:

Let points C and D be located on a semicircle with diameter AB. Let E be the intersection of AD and BC and F the foot of the perpendicular from E to AB. Then EF is the bisector of angle CFD.

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Note that since inscribed angles ACB and ADB subtend a diameter they are right. So that AD and BC are altitudes in ΔABS where S is the intersection of AC and BD, E is the orthocenter of this triangle and EF is the third altitude. What is claimed is nothing more nor less than the mirror property of the altitude. Greg's shows how the latter can be derived from Pascal's theorem of the hexagram.

### Proof

Reflect C and D in AB to obtain C' and B'. Consider the hexagon ADC'BCD'. By Pascal's theorem, the three points at which the pairs

### References

- G. Markowsky,
*Pascal's Hexagon Theorem implies the Butterfly Theorem*, Mathematics Magazine, Volume 84, Number 1, February 2011 , pp. 56-62(7)

### Pascal and Brianchon Theorems

- Pascal's Theorem
- Pascal in Ellipse
- Pascal's Theorem, Homogeneous Coordinates
- Projective Proof of Pascal's Theorem
- Pascal Lines: Steiner and Kirkman Theorems
- Brianchon's theorem
- Brianchon in Ellipse
- The Mirror Property of Altitudes via Pascal's Hexagram
- Pappus' Theorem
- Pencils of Cubics
- Three Tangents, Three Chords in Ellipse
- MacLaurin's Construction of Conics
- Pascal in a Cyclic Quadrilateral
- Parallel Chords
- Parallel Chords in Ellipse
- Construction of Conics from Pascal's Theorem
- Pascal: Necessary and Sufficient
- Diameters and Chords
- Chasing Angles in Pascal's Hexagon
- Two Triangles Inscribed in a Conic
- Two Triangles Inscribed in a Conic - with Solution
- Two Pascals Merge into One
- Surprise: Right Angle in Circle

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Copyright © 1996-2017 Alexander Bogomolny61209930 |