Question

如图 解题？

Answer 1

解：如图 1，以 BC 为一边构造等边 △BCF，连接DF、AF.
∠BAD = ∠BDA = 70° ⇒ ∠ABD = 40°，∠BAC = ∠ABC = 70° ⇒ ∠ACB = 40°.
∴ ∠DBC = ∠ABC - ∠ABD = 70° - 40° = 30°，
∴ ∠DBF = ∠CBF - ∠DBC = 60° - 30° = 30° = ∠DBC，
又 ∵ BD = BD，BF = BC，
∴ △FDB ≌ △CDB (SAS).
∴ ∠DFB = ∠DCB = 40°，DF = DC .
DF = DC ⇒ ∠DFC = ∠DCF = ∠BCF - ∠BCA = 60° - 40° = 20°.
∴∠FDA =∠DFC + ∠DCF = 20° + 20° = 40°.
∠BAC = ∠ABC = 70°⇒AC = BC，
又∵ FC = BC，
∴ FC = BC = AC.
∴ ∠CFA =∠CAF = (180°-∠ACF/2) = (180°-20°/2) = 80°.
AC = BC ，CD = BE. ⇒ CE = DA，
又 ∵∠DCE = 40° = ∠FDA，CD = DF .
∴ △CDE ≌ △DFA(SAS).
∴ ∠CDE = ∠DFA = ∠CFA - ∠CFD = 80° - 20° = 60°.
∴ ∠BDE = 180° - ∠CDE - ∠ADB = 180° - 60° - 70° = 50°.

Answer 2

这道题从图上看有abcd4个角，这个你最好去问问你的老师

Answer 3

你们班长城路由器。