如圖,設(shè)
C為線段
AB的中點(diǎn),
BCDE是以
BC為一邊的正方形,以
B為圓心,
BD為半徑的圓與
AB及其延長(zhǎng)線相交于點(diǎn)
H及
K.
(Ⅰ)求證:
HC·
CK=
BC2;
(Ⅱ)若圓的半徑等于2,求
AH·
AK的值.
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(Ⅰ)連結(jié)
DH,
DK,則
DH⊥
DK,
∴△
DHC∽△
KDC,∴
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,
DC2=
HC·
CK,
又
DC=
BC,∴
BC2=
HC·
CK………………(5分)
(Ⅱ)連結(jié)
AD,則
AD⊥
BD,
AD=
BD,∴
AD是⊙
B的切線,于是
AD2=
AH·
AK,
∴
AH·
AK=4
(I)證明可以從結(jié)論出發(fā)進(jìn)行尋找解題途徑
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.
(II)證明AD為圓的切線之后,利用切割線定理即可求解
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