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In matematica , e in particolare in geometria euclidea , il teorema della farfalla afferma che:
sia
M
{\displaystyle M}
punto medio di una corda
P
Q
{\displaystyle PQ}
cerchio e siano
A
B
{\displaystyle AB}
C
D
{\displaystyle CD}
M
{\displaystyle M}
X
{\displaystyle X}
Y
{\displaystyle Y}
A
D
{\displaystyle AD}
B
C
{\displaystyle BC}
P
Q
{\displaystyle PQ}
M
{\displaystyle M}
X
Y
{\displaystyle XY}
Dimostrazione del Teorema della farfalla Siano
X
X
′
{\displaystyle XX'}
X
X
″
{\displaystyle XX''}
X
{\displaystyle X}
A
M
{\displaystyle AM}
D
M
{\displaystyle DM}
Y
Y
′
{\displaystyle YY'}
Y
Y
″
{\displaystyle YY''}
Y
{\displaystyle Y}
B
M
{\displaystyle BM}
C
M
{\displaystyle CM}
Adesso, poiché
△
M
X
X
′
∼
△
M
Y
Y
′
,
{\displaystyle \triangle MXX'\sim \triangle MYY',\,}
M
X
M
Y
=
X
X
′
Y
Y
′
,
{\displaystyle {MX \over MY}={XX' \over YY'},}
△
M
X
X
″
∼
△
M
Y
Y
″
,
{\displaystyle \triangle MXX''\sim \triangle MYY'',\,}
M
X
M
Y
=
X
X
″
Y
Y
″
,
{\displaystyle {MX \over MY}={XX'' \over YY''},}
△
A
X
X
′
∼
△
C
Y
Y
″
,
{\displaystyle \triangle AXX'\sim \triangle CYY'',\,}
X
X
′
Y
Y
″
=
A
X
C
Y
,
{\displaystyle {XX' \over YY''}={AX \over CY},}
△
D
X
X
″
∼
△
B
Y
Y
′
,
{\displaystyle \triangle DXX''\sim \triangle BYY',\,}
X
X
″
Y
Y
′
=
D
X
B
Y
,
{\displaystyle {XX'' \over YY'}={DX \over BY},}
Dalle precedenti equazioni, si può facilmente dedurre che
(
M
X
M
Y
)
2
=
X
X
′
Y
Y
′
X
X
″
Y
Y
″
,
{\displaystyle \left({MX \over MY}\right)^{2}={XX' \over YY'}{XX'' \over YY''},}
=
A
X
.
D
X
C
Y
.
B
Y
,
{\displaystyle {}={AX.DX \over CY.BY},}
=
P
X
.
Q
X
P
Y
.
Q
Y
,
{\displaystyle {}={PX.QX \over PY.QY},}
=
(
P
M
−
X
M
)
.
(
M
Q
X
M
)
(
P
M
M
Y
)
.
(
Q
M
−
M
Y
)
,
{\displaystyle {}={(PM-XM).(MQ XM) \over (PM MY).(QM-MY)},}
=
(
P
M
)
2
−
(
M
X
)
2
(
P
M
)
2
−
(
M
Y
)
2
,
{\displaystyle {}={(PM)^{2}-(MX)^{2} \over (PM)^{2}-(MY)^{2}},}
poiché
P
M
{\displaystyle PM}
M
Q
{\displaystyle MQ}
Ora,
(
M
X
)
2
(
M
Y
)
2
=
(
P
M
)
2
−
(
M
X
)
2
(
P
M
)
2
−
(
M
Y
)
2
.
{\displaystyle {(MX)^{2} \over (MY)^{2}}={(PM)^{2}-(MX)^{2} \over (PM)^{2}-(MY)^{2}}.}
Pertanto, possiamo concludere che
M
X
=
M
Y
,
{\displaystyle MX=MY,}
M
{\displaystyle M}
X
Y
.
{\displaystyle XY.}
H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967.
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