## Two Triangles With Common Base and Altitude

What is this about?

A Mathematical Droodle

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener| |Store|

Copyright © 1996-2017 Alexander BogomolnyThe applet may suggest a statement by Greg Markowsky. If C and D are two points on a semicircle with diameter AB, E is the intersection of AD and BC and F the foot of perpendicular from E to AB, then Greg calls C and D a *pair of reflection* about F based on an observation of his.

Greg's statement admits a reformulation: Given two triangles ABS and ABT with altitudes AD, BC, FS and AV, BU, FT, respectively. Then DU and CV intersect on the common altitude through F.

What if applet does not run? |

### Proof

Consider the hexagon ADUBCV and a perpendicular FG to AB. By the mirror property of altitudes AD and BC cross on FG and so are VA and UB. Let X be the intersection of DU and CV. By Pascal's Hexagram Theorem, the three points of intersection are collinear. Since the first two lie on FG, so does X.

### References

- G. Markowsky,
*Pascal’s Hexagon Theorem implies the Butterfly Theorem*, 2007, submitted for publication

|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener| |Store|

Copyright © 1996-2017 Alexander Bogomolny62104078 |