BD、CD分别是△ABC的两个外角∠CBE、∠BCF的平分线,试探索∠BDC与∠A之间的数量关系.

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BD、CD分别是△ABC的两个外角∠CBE、∠BCF的平分线,试探索∠BDC与∠A之间的数量关系.

BD、CD分别是△ABC的两个外角∠CBE、∠BCF的平分线,试探索∠BDC与∠A之间的数量关系.
BD、CD分别是△ABC的两个外角∠CBE、∠BCF的平分线,试探索∠BDC与∠A之间的数量关系.

BD、CD分别是△ABC的两个外角∠CBE、∠BCF的平分线,试探索∠BDC与∠A之间的数量关系.
∵∠A+∠ABC+∠ACB=180
∴∠ABC+∠ACB=180-∠A
∵∠EBC=180-∠ABC,BD平分∠EBC
∴∠DBC=∠EBC/2=90-∠ABC/2
∵∠FCB=180-∠ACB,CD平分∠FCB
∴∠DCB=∠FCB/2=90-∠ACB/2
∵∠BDC+∠DBC+∠DCB=180
∴∠BDC=180-(∠DBC+∠DCB)
=180-(90-∠ABC/2+90-∠ACB/2)
=(∠ABC+∠ACB)/2
=(180-∠A)/2
=90-∠A/2

∠BDC=180-∠CBD-∠BCD
=180-1/2∠CBE-1/2∠BCF
=180-1/2(∠A+∠ACB)-1/2(∠A+∠ABC)
=180-1/2∠A-1/2∠ACB-1/2∠A-∠ABC
=180-∠A-1/2...

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∠BDC=180-∠CBD-∠BCD
=180-1/2∠CBE-1/2∠BCF
=180-1/2(∠A+∠ACB)-1/2(∠A+∠ABC)
=180-1/2∠A-1/2∠ACB-1/2∠A-∠ABC
=180-∠A-1/2(∠ACB+∠ABC)
=180-∠A-1/2(180-∠A)
=180-∠A-90+1/2∠A
=90-1/2∠A

收起

∠BDC=180-1/2(∠CBE+∠BCF)
=180°-1/2(2∠A+∠ABC+∠ACB)
=180°-1/2*∠A-1/2(∠A+∠ABC+∠ACB)
=180°-1/2*∠A-1/2*180°
=180°-1/2∠A-90°
=90°-∠A/2