(20 14? As shown in the figure, the square ABCD and the square BEFG have a common vertex B, the point G is on the BC side, and the extension line of AG intersects with CE at the point H to connect BH.

(1) proves that ∵ quadrilateral ABCD and BEFG are both squares,

∴AB=CB,∠ABC=∠CBE=90，GB=EB，

In △ABG and △ BC,

∫AB = CB∠ABG =∠CBE GB = EB，

∴△ABG≌△BCE(SAS)，

∴∠bag=∠bce；

(2) the connection communication,

∫from( 1):∠BAG =∠BCE，

∴∠bag+∠beh=∠bce+∠beh= 180-∠CBE = 90，

∴∠ahe= 180-(∠ bag+∠ shell) =90

∠∠AGB =∠CGH，

∴△AGB∽△CGH，

∴AGCG=BGHG，

∴HGCG=BGAG，

∫∠BGH =∠AGC，

∴△BGH∽△AGC，

∴BHAC=BGAG，

Namely BH? AG=AC？ BG，

At Rt△AHE and 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) From (2): BHAH=AC? BGAB(AB+BE)，

AB = kBG，

∴∴BHAH=