## Two Triangles With Common Base and Altitude

What is this about?

A Mathematical Droodle

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Copyright © 1996-2018 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.

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### 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|

Copyright © 1996-2018 Alexander Bogomolny65120620 |