如圖,△ABC中,AB=AC,∠BAC=90°,AE=
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AC,BD=
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AB,點(diǎn)F在BC上,且CF=
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BC.求證:
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(1)EF⊥BC;
(2)∠ADE=∠EBC.
證明:設(shè)AB=AC=3a,
則AE=BD=a,CF=
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a.
(1)
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=
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=
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,

=
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=
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.
又∠C為公共角,故△BAC∽△EFC,
由∠BAC=90°.∴∠EFC=90°,∴EF⊥BC.
(2)由(1)得EF=
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a,
故
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=
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=
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,

=

=
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,
∴
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=
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.∵∠DAE=∠BFE=90°,
∴△ADE∽△FBE,∴∠ADE=∠EBC.
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