课堂点睛数学七年级下册第34页15题答案
∵∠BAC=180°-(∠ABC ∠ACB) ∴1/2∠BAC=90°-1/2(∠ABC ∠ACB) ∴1/2(∠ABC ∠ACB)=90°-1/2∠BAC ∵∠BPC=180°-(∠PBC ∠PCB) ∠DBC=(180°-∠ABC)÷2=90°-1/2∠ABC ∠DCB=(180°-∠ACB)÷2=90°-1/2∠ACB ∴∠DBC ∠DCB=∠180°-1/2(∠ABC ∠ACB) ∴∠BDC=1/2(∠ABC ∠ACB) ∴∠BDC=90°-1/2∠BAC...