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。