如图,∠ABC=∠ACB,AD、BD、CD分别平分△ABC的外角∠EAC、内角∠ABC、外角∠ACF.以下结论:
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如图,∠ABC=∠ACB,AD、BD、CD分别平分△ABC的外角∠EAC、内角∠ABC、外角∠ACF.以下结论:
①AD∥BC;
②∠ACB=2∠ADB;
③∠ADC=90°-∠ABD;
④BD平分∠ADC;
⑤∠BDC=
①AD∥BC;
②∠ACB=2∠ADB;
③∠ADC=90°-∠ABD;
④BD平分∠ADC;
⑤∠BDC=
1 |
2 |
∵AD平分∠EAC,
∴∠EAC=2∠EAD,
∵∠EAC=∠ABC+∠ACB,∠ABC=∠ACB,
∴∠EAD=∠ABC,
∴AD∥BC,∴①正确;
∵AD∥BC,
∴∠ADB=∠DBC,
∵BD平分∠ABC,∠ABC=∠ACB,
∴∠ABC=∠ACB=2∠DBC,
∴∠ACB=2∠ADB,∴②正确;
∵AD平分∠EAC,CD平分∠ACF,
∴∠DAC=
1
2∠EAC,∠DCA=
1
2∠ACF,
∵∠EAC=∠ACB+∠ACB,∠ACF=∠ABC+∠BAC,∠ABC+∠ACB+∠BAC=180°,
∴∠ADC=180°-(∠DAC+∠ACD)
=180°-
1
2(∠EAC+∠ACF)
=180°-
1
2(∠ABC+∠ACB+∠ABC+∠BAC)
=180°-
1
2(180°-∠ABC)
=90°-
1
2∠ABC,∴③正确;
∵BD平分∠ABC,
∴∠ABD=∠DBC,
∵∠ADB=∠DBC,∠ADC=90°-
1
2∠ABC,
∴∠ADB不等于∠CDB,∴④错误;
∵∠ACF=2∠DCF,∠ACF=∠BAC+∠ABC,∠ABC=2∠DBC,∠DCF=∠DBC+∠BDC,
∴∠BAC=2∠BDC,∴⑤正确;
即正确的有4个,
故选B.
∴∠EAC=2∠EAD,
∵∠EAC=∠ABC+∠ACB,∠ABC=∠ACB,
∴∠EAD=∠ABC,
∴AD∥BC,∴①正确;
∵AD∥BC,
∴∠ADB=∠DBC,
∵BD平分∠ABC,∠ABC=∠ACB,
∴∠ABC=∠ACB=2∠DBC,
∴∠ACB=2∠ADB,∴②正确;
∵AD平分∠EAC,CD平分∠ACF,
∴∠DAC=
1
2∠EAC,∠DCA=
1
2∠ACF,
∵∠EAC=∠ACB+∠ACB,∠ACF=∠ABC+∠BAC,∠ABC+∠ACB+∠BAC=180°,
∴∠ADC=180°-(∠DAC+∠ACD)
=180°-
1
2(∠EAC+∠ACF)
=180°-
1
2(∠ABC+∠ACB+∠ABC+∠BAC)
=180°-
1
2(180°-∠ABC)
=90°-
1
2∠ABC,∴③正确;
∵BD平分∠ABC,
∴∠ABD=∠DBC,
∵∠ADB=∠DBC,∠ADC=90°-
1
2∠ABC,
∴∠ADB不等于∠CDB,∴④错误;
∵∠ACF=2∠DCF,∠ACF=∠BAC+∠ABC,∠ABC=2∠DBC,∠DCF=∠DBC+∠BDC,
∴∠BAC=2∠BDC,∴⑤正确;
即正确的有4个,
故选B.
如图,∠ABC=∠ACB,AD、BD、CD分别平分△ABC的外角∠EAC、内角∠ABC、外角∠ACF.以下结论:
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