已知△ABC,作等腰△ABD与等腰△ACE,使AB=AD,AC=AE,∠BAD=∠CAE,直线CD、BE交于点O.
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已知△ABC,作等腰△ABD与等腰△ACE,使AB=AD,AC=AE,∠BAD=∠CAE,直线CD、BE交于点O.
若 ∠BAD=∠CAE=α,α为锐角,连接AO,则∠AOE=?
如图:
若 ∠BAD=∠CAE=α,α为锐角,连接AO,则∠AOE=?
如图:
过点A作AM⊥BE于M,AN⊥CD于N
∵∠BAE=∠BAC+∠CAE,∠DAC=∠BAC+∠BAD,∠BAD=∠CAE=α
∴∠BAE=∠DAC
∵AB=AD,AC=AE
∴△ABE≌△ADC (SAS)
∴BE=CD,S△ABE=S△ADC,∠ABE=∠ADC
∴∠DOE=∠DBE+∠BDC
=∠ABD+∠ABE+∠BDC
=∠ABD+∠ADC+∠BDC
=∠ABD+∠ADB
=180-∠BAD
=180-α
∵AM⊥BE,AN⊥CD
∴S△ABE=BE×AM/2,S△ADC=CD×AN/2
∴BE×AM/2=CD×AN/2
∴AM=AN
∴OA平分∠DOE
∴∠AOE=∠DOE/2=(180-α)/2
数学辅导团解答了你的提问,
∵∠BAE=∠BAC+∠CAE,∠DAC=∠BAC+∠BAD,∠BAD=∠CAE=α
∴∠BAE=∠DAC
∵AB=AD,AC=AE
∴△ABE≌△ADC (SAS)
∴BE=CD,S△ABE=S△ADC,∠ABE=∠ADC
∴∠DOE=∠DBE+∠BDC
=∠ABD+∠ABE+∠BDC
=∠ABD+∠ADC+∠BDC
=∠ABD+∠ADB
=180-∠BAD
=180-α
∵AM⊥BE,AN⊥CD
∴S△ABE=BE×AM/2,S△ADC=CD×AN/2
∴BE×AM/2=CD×AN/2
∴AM=AN
∴OA平分∠DOE
∴∠AOE=∠DOE/2=(180-α)/2
数学辅导团解答了你的提问,
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