如图,△ABC的三条角平分线交于点O,过点O作OE⊥BC于点E,求证:∠BOD=∠COE.
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如图,△ABC的三条角平分线交于点O,过点O作OE⊥BC于点E,求证:∠BOD=∠COE.
证明:∵∠AFO=∠FBC+∠ACB=
1
2∠ABC+∠ACB,
∴∠AOF=180°-(∠DAC+∠AF0)
=180°-[
1
2∠BAC+
1
2∠ABC+∠ACB]
=180°-[
1
2(∠BAC+∠ABC)+∠ACB]
=180°-[
1
2(180°-∠ACB)+∠ACB]
=180°-[90°+
1
2∠ACB]
=90°-
1
2∠ACB,
∴∠BOD=∠AOF=90°-
1
2∠ACB,
又∵在直角△OCE中,∠COE=90°-∠OCD=90°-
1
2∠ACB,
∴∠BOD=∠COE.
1
2∠ABC+∠ACB,
∴∠AOF=180°-(∠DAC+∠AF0)
=180°-[
1
2∠BAC+
1
2∠ABC+∠ACB]
=180°-[
1
2(∠BAC+∠ABC)+∠ACB]
=180°-[
1
2(180°-∠ACB)+∠ACB]
=180°-[90°+
1
2∠ACB]
=90°-
1
2∠ACB,
∴∠BOD=∠AOF=90°-
1
2∠ACB,
又∵在直角△OCE中,∠COE=90°-∠OCD=90°-
1
2∠ACB,
∴∠BOD=∠COE.
如图,△ABC的三条角平分线交于点O,过点O作OE⊥BC于点E,求证:∠BOD=∠COE.
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