如图,△ABC中,∠A=40度,AB=AC,D为△ABC内一点,且∠DCA=∠DBC,求∠BDC的度数.

问题描述:

如图,△ABC中,∠A=40度,AB=AC,D为△ABC内一点,且∠DCA=∠DBC,求∠BDC的度数.

延长CD交AB于E
∵AB=AC,∠A=40
∴∠ABC=∠ACB=(180-∠A)/2=70
∵∠BEC=∠A+∠DCA,∠BDC=∠BEC+∠DBA,∠DCA=∠DBC
∴∠BDC=∠A+∠DCA+∠DBA
=∠A+∠DBC+∠DBA
=∠A+∠ABC
=40+70
=110°