Difference between revisions of "Butterfly Theorem"
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==Proof== | ==Proof== | ||
− | This simple proof uses projective geometry. | + | This simple proof uses [[projective geometry]]. |
+ | |||
First we note that <math>(AP, AB; AD, AQ) = (CP, CB; CD, CQ).</math> | First we note that <math>(AP, AB; AD, AQ) = (CP, CB; CD, CQ).</math> | ||
Therefore, | Therefore, | ||
Line 15: | Line 16: | ||
<math>\blacksquare</math>. | <math>\blacksquare</math>. | ||
− | + | ==Related Reading== | |
+ | http://agutie.homestead.com/FiLEs/GeometryButterfly.html | ||
− | http:// | + | http://www.mathematik.uni-muenchen.de/~fritsch/butterfly.pdf |
==See also== | ==See also== |
Latest revision as of 12:32, 26 January 2015
Let be the midpoint of chord
of a circle, through which two other chords
and
are drawn.
and
intersect chord
at
and
, respectively. The Butterfly Theorem states that
is the midpoint of
.
Proof
This simple proof uses projective geometry.
First we note that
Therefore,
Since
,
Moreover,
so
as desired.
.
Related Reading
http://agutie.homestead.com/FiLEs/GeometryButterfly.html
http://www.mathematik.uni-muenchen.de/~fritsch/butterfly.pdf