∴AB=CB,∠ABC=∠CBE=90,GB=EB,
在△ABG和△公元前,
∫AB = CB∠ABG =∠CBE GB = EB,
∴△ABG≌△BCE(SAS),
∴∠bag=∠bce;
(2)連接交流、
∫from(1):∠BAG =∠BCE,
∴∠bag+∠beh=∠bce+∠beh=180-∠CBE = 90,
∴∠ahe=180-(∠包+∠貝)=90
∠∠AGB =∠CGH,
∴△AGB∽△CGH,
∴AGCG=BGHG,
∴HGCG=BGAG,
∫∠BGH =∠AGC,
∴△BGH∽△AGC,
∴BHAC=BGAG,
即BH?AG=AC?BG,
在Rt△AHE和Rt△ABG,
* cosHAE = AHAE = ABAG,
∴AH?AG=AB?AE,
∴BH?AGAH?AG=AC?BGAB?AE,
∴BHAH=AC?BGAB(AB+BE),
AB = 2BG,
∴BHAH=2AB?BGAB?3BG = 23
(3)從(2): BHAH=AC?BGAB(AB+BE),
AB = kBG,
∴∴BHAH=