∠EAC=∠DAB,AF=AC,AD=AB,BC的延长线交AD于点F,交AE延长线于点G,∠ACB=105°,∠CAD=10°,∠ADE=25°,求∠DFB和∠AGB的度数
人气:184 ℃ 时间:2020-08-28 09:49:41
解答
∵∠acb=105°
∴∠ace=75°
∵∠cad=10°
∴∠afc=95° ∠dfb=180°-95°=85°
∵ea=ca(已知)
ab=ad(已知)
∠cab=∠dae(∠eac=∠dab,∠dac为公共角)
∴△cab≌△dea
∠acb=∠dea=105°
∠geo=75°(设de与gb相交的点为o)
∠bge=180°-75°-60°=45°
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