数学图形题!急!!!
问题见图:
作DE⊥AB于E,DF⊥BC于F,DG⊥AC于G
易证DE=DF=DG
∴AD平分∠BAC
∠DBC=1/2 ∠EBC=1/2(∠A+∠ACB)
∠DCB=1/2 ∠GCB=1/2(∠A+∠ABC)
∠BDC=180º-∠DBC-∠DCB=180º-1/2(∠A+∠ACB)-1/2(∠A+∠ABC)
=180º-1/2∠A-1/2∠ACB-1/2∠A-1/2∠ABC
∵1/2∠A+1/2∠ACB+1/2∠ABC=90º
∴∠BDC=90º-1/2∠A
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