本文以课本基本图形(如图1、图2)为依托,对一道中考题进行多解与变式如下.题目(2013年绍兴)在△ABC中,∠CAB=90°,AD⊥BC于点D,点E为AB的中点,EC与AD交于点G,点F在BC上.(1)如图3,AC∶AB=1∶2,EF⊥CB,求证:EF=CD;(2)如图4,AC∶AB=1∶31/2,EF⊥CE,求EF∶EG的值. The text (Figure 1, Figure 2) as the basis, a multi-medium test questions and variants of the following questions. (2013 Shaoxing) △ ABC, ∠ CAB = 90 °, AD ⊥ BC at the point D, point E is the midpoint of AB, EC and AD intersect at point G, and point F is at BC (1) As shown in Figure 3, AC: AB = 1: 2, EF⊥CB, 2) As shown in Figure 4, AC: AB = 1:31 / 2, EF⊥CE, EF: EG value.