题目内容
已知:如图,A是⊙O1、⊙O2的一个交点,点M是O1O2的中点,过点A的直线BC垂直于MA,分别交⊙O1、⊙O2于B、C.
(1)求证:AB=AC;
(2)若O1A切⊙O2于点A,弦AB、AC的弦心距分别为dl、d2,求证:d1+d2=O1O2;
(3)在(2)条件下,若d1d2=1,设⊙O1、⊙O2的半径分别为R、r,求证:R2+r2=
.
(1)求证:AB=AC;
(2)若O1A切⊙O2于点A,弦AB、AC的弦心距分别为dl、d2,求证:d1+d2=O1O2;
(3)在(2)条件下,若d1d2=1,设⊙O1、⊙O2的半径分别为R、r,求证:R2+r2=
(R2+r2)2 |
R2r2 |
证明:(1)分别作O1D⊥AB于点D,O2E⊥AC于点E.
则AB=2AD,AC=2AE.
∵O1D∥AM∥O2E,
∵M为O1O2的中点,
∴AD=AE,AB=AC.
(2)∵O1A切⊙O2于点A,
∴O1A⊥O2A,
又∵M为O1O2的中点,O1O2=2AM
在梯形O1O2ED中,
∵AM为梯形的中位线,O1D+O2E=2AM,
∴O1D+O2E=O1O2,
即d1+d2=O1O2.
(3)∵O1A⊥O2A,
∴∠AO1D=∠O2AE,
∴Rt△O1AD∽Rt△AO2E.
∴
=
=
,
即
=
=
.
∴AD•AE=d1•d2=1.
即由(1)(2)知,AD=AE=1,O1O2=d1+d2,
∴d1=
,d2=
,
∴R2+r2=O1O22=(d1+d2)2=(
+
)2=
.
则AB=2AD,AC=2AE.
∵O1D∥AM∥O2E,
∵M为O1O2的中点,
∴AD=AE,AB=AC.
(2)∵O1A切⊙O2于点A,
∴O1A⊥O2A,
又∵M为O1O2的中点,O1O2=2AM
在梯形O1O2ED中,
∵AM为梯形的中位线,O1D+O2E=2AM,
∴O1D+O2E=O1O2,
即d1+d2=O1O2.
(3)∵O1A⊥O2A,
∴∠AO1D=∠O2AE,
∴Rt△O1AD∽Rt△AO2E.
∴
AD |
O2E |
O1D |
AE |
O1A |
O2A |
即
AD |
d2 |
d1 |
AE |
R |
r |
∴AD•AE=d1•d2=1.
即由(1)(2)知,AD=AE=1,O1O2=d1+d2,
∴d1=
R |
r |
r |
R |
∴R2+r2=O1O22=(d1+d2)2=(
R |
r |
r |
R |
(R2+r2)2 |
R2r2 |
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