(2008•淄博)正方形ABCD的對角線交點(diǎn)為O,兩條對角線把它分成了四個(gè)面積相等的三角形.
(1)平行四邊形ABCD的兩條對角線交點(diǎn)為O,若△AOB,△BOC,△COD,△DOA面積分別為S1,S2,S3,S4,試判斷S1,S2,S3,S4的關(guān)系,并加以證明;
(2)四邊形ABCD的兩條對角線互相垂直,交點(diǎn)為O,若△AOB,△BOC,△COD,△DOA面積分別為S1,S2,S3,S4,試判斷S1,S2,S3,S4的關(guān)系,并加以證明;
(3)四邊形ABCD的兩條對角線交點(diǎn)為O,若△AOB,△BOC,△COD,△DOA面積分別為S1,S2,S3,S4,試判斷S1,S2,S3,S4的關(guān)系,并加以證明;
(4)四邊形ABCD的兩條對角線相等,交點(diǎn)為O,∠BAC=∠BDC,若△AOB,△BOC,△COD,△DOA面積分別為S1,S2,S3,S4,試只用S1,S3或只用S2,S4表示四邊形ABCD的面積S.
【答案】
分析:(1)根據(jù)平行四邊形的性質(zhì)可證得四個(gè)小三角形面積相等;
(2)我們可以表示出這四個(gè)面積,S
1=
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OA•OB,S
2=
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OB•OC,S
3=
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OC•OD,S
4=
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OD•OA,于是我們發(fā)現(xiàn)S
1S
3=S
2S
4;
(3)雖然兩條對角線不垂直了,但是思路和(2)是一樣的;
(4)應(yīng)該分AB與CD平行或不平行兩種情況進(jìn)行分析.
解答:
解:(1)∵四邊形ABCD是平行四邊形,
∴OA=OC,
∵△AOB,△BOC的邊OA,OC上的高相同,
∴S
1=S
2,
同理S
2=S
3,S
3=S
4,S
4=S
1,
∴S
1=S
2=S
3=S
4.
(2)∵AC⊥BD,垂足為O,
∴S
1=

OA•OB,S
2=
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OB•OC,S
3=
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OC•OD,S
4=
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OD•OA,
∴S
1S
3=S
2S
4;
(3)設(shè)點(diǎn)B到線段AC所在直線的距離為h
1,點(diǎn)D到線段AC所在直線的距離為h
2,
∴S
1=

OA•h
1,S
2=
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OC•h
1,S
3=
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OC•h
2,S
4=
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OA•h
2,
∴S
1S
3=S
2S
4;
(4)∵BAC=∠BDC,∠AOB=∠DOC,
∴∠DCA=∠ABD,
當(dāng)AB與CD不平行時(shí),必相交于一點(diǎn),
設(shè)線段BA與CD的延長線交于點(diǎn)E,
∵AC=BD,∠AEC=∠DEB,
∴△AEC≌△DEB,
∴AE=DE,CE=BE,
∴AB=DC,
∴△AOB≌△DOC,
∴S
1=S
3,
∵S
1S
3=S
2S
4,
∴S
12=S
2S
4,
∴S=S
1+S
2+S
3+S
4=2S
1+S
2+S
4=S
2+S
4+2

(或=(
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+
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)
2);
當(dāng)AB與CD平行時(shí),則△ABD與△BAC同底等高,有S
1+S
2=S
1+S
4,
∴S
2=S
4,
∵S
1S
3=S
2S
4,
∴S
22=S
1S
3,S=S
1+S
3+2S
2=S
1+S
3+2
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(或=(
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+
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)
2).
點(diǎn)評:本題主要考查了全等三角形的性質(zhì)以及三角形面積公式的靈活運(yùn)用.要注意(4)中要分AB,CD平行和不平行兩種情況來求解.