证明∠D=90°+2分之1∠A,如下:
∠D=180°-(∠DBC+∠DCB)
=180°-(∠ABC/2+∠ACB/2)
=180°-(∠ABC+∠ACB)/2
=180°-(180°-∠A)/2=90°+∠A/2
(1)(2)仅给出结论,证明不予给出!
(1)∠D=∠A/2
(2)∠D=90°-∠A/2
证明∠D=90°+2分之1∠A,如下:
∠D=180°-(∠DBC+∠DCB)
=180°-(∠ABC/2+∠ACB/2)
=180°-(∠ABC+∠ACB)/2
=180°-(180°-∠A)/2=90°+∠A/2
(1)(2)仅给出结论,证明不予给出!
(1)∠D=∠A/2
(2)∠D=90°-∠A/2