Simone课堂点睛数学七年言赵海概气级下册第34页15题答案
的有关信息介绍如下：
∵∠BAC=180°-(∠ABC+∠ACB)
∴1来自/2∠BAC=90°-1/2(∠ABC+∠ACB)
∴1/2(∠ABC+∠ACB)=90°-1/2∠BAC
∵∠BPC=180°-(∠PBC+∠PCB)
∠DBC=(180°按似精-∠ABC)÷2=90°-1/360问答2∠ABC
∠DCB=(180°-∠ACB)÷2=90°-1/2∠ACB
∴∠DBC+∠DCB=∠180°-1/2(∠ABC+∠ACB)
∴∠BDC=1/2(∠ABC+∠ACB)
∴∠BDC=90又国军军两的球导小主措°-1/2∠BAC
