两道数学求证（追加20）

∵∠A+∠ABC+∠ACB=180°
∴∠ABC+∠ACB=180°-∠A
又∵BD、CD平缺哪分闹扮慧∠ABC、∠ACB
∴∠DBC+∠DCB=∠ABD+∠ACD=(∠ABC+∠ACB)/2
∵∠DBC+∠液答BCD+∠CDB=180°
∴∠D=180°-(∠DBC+∠BCD)=180-(∠ABC+∠ACB)/2
=180°-(180°-∠A)/2=180°-90°+∠A/2
=90°+∠A/2

1、∵∠D+1/2∠B+1/脊咐好2∠C=180° ⑴
∠A+∠B+∠C=180° ⑵
⑴×2-⑵得：2∠D+180°-∠A=360°
∴∠D=90°+1/2∠樱铅A
2、∵∠F+1/2∠CBE+1/2∠BCD=180° ⑴
∠A+(180°-∠CBE)+(180°-∠BCD)=180°⑵
⑴×2+⑵简念得：2∠F+∠A+360=540°
∴∠F=90°-1/2∠A