Difference between revisions of "Butterfly Theorem"
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==Proof== | ==Proof== | ||
− | {{ | + | This simple proof uses projective geometry. |
− | + | First we note that <math>(AP, AB; AD, AQ) = (CP, CB; CD, CQ).</math> | |
− | + | Therefore, | |
+ | <cmath>\frac{(PX)(MQ)}{(PQ)(MX)} = \frac{(PM)(YQ)}{(PQ)(YM)}.</cmath> | ||
+ | Since <math>MQ = PM</math>, | ||
+ | <cmath>\frac{MX}{YM} = \frac{XP}{QY}.</cmath> | ||
+ | Moreover, | ||
+ | <cmath>\frac{MX + PX}{YM + QY} = 1,</cmath> | ||
+ | so <math>MX = YM,</math> as desired. | ||
+ | <math>\blacksquare</math>. | ||
Link to a good proof : | Link to a good proof : | ||
http://agutie.homestead.com/FiLEs/GeometryButterfly.html | http://agutie.homestead.com/FiLEs/GeometryButterfly.html | ||
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==See also== | ==See also== |
Revision as of 18:30, 31 May 2011
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.
.
Link to a good proof :
http://agutie.homestead.com/FiLEs/GeometryButterfly.html