Question

如图,已知点D为△ABC内一点,求证:∠BDC>∠A

急求,,,,

Answer 1

如果∠1是∠ABD,∠2是∠ACD.

∵∠1+∠DBC+∠2+∠BCD=180﹣∠A

又∵∠DBC+∠BCD+∠BDC=180

∴∠1﹢∠2＝∠BDC﹣∠A
∴∠BDC>∠A

Answer 2

因为∠A+∠ABC+∠ACB=180°
∠BDC+∠CBD+∠BCD=180°
所以∠A=180°-(∠ABC+∠ACB)
∠BDC=180°-(∠CBD+∠BCD)
又因为∠CBD<∠ABC,∠BCD<∠ACB
所以∠CBD+∠BCD<∠ABC+∠ACB
于是180°-(∠CBD+∠BCD)>180°-(∠ABC+∠ACB)
即∠BDC＞∠A