Question

请高手帮忙解出第三个问题

Answer 1

⑶ ∠DGF＝45°.提示：①作BH⊥FP于H则∠BPF＝90°－∠BCD＝90°－∠BAC＝∠ADE∴△HBP∽△EAD∽△EFB∽△HFB [其中△EAD∽△EFB理由“SAS”,而∠PBF＝﹙180°－∠BAC﹚－∠ABF＝180°－∠BAC－∠ADE＝90°∴∠HBF＝90°－∠PBH＝90°－∠BAC＝90°－∠BFE＝∠EBF∴△EFB≌△HFB]∴BP＝2/1.5·BF＝2/1.5×2.5＝10/3,PC＝5/3,AG/GC＝AD/PC＝3,

②连接BD交AC于I，则BD⊥AC，AI＝IC,由①AG＝3GC∴IG＝GC＝√5＝DI∴∠AGD＝45°

﹙其中△AID∽△AEF∴ID/EF＝IA/AE＝AD/AF＝5/﹙3√5/2﹚∴AC＝2AI＝4√5,DI＝√5＝CG﹚