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原题再现:(2015年中考数学安徽卷)如图1,在四边形ABCD中,点E,F分别是AB,CD的中点,过点E作AB的垂线,过点F作CD的垂线,两垂线交于点G,联结AG,BG,CG,DG,且∠AGD=∠BGC。(1)求证:AD=BC;(2)求证:△AGD∽△EGF;(3)如图2,若AD、BC所在直线互相垂直,求(AD/EF)的值。(1)证明:因为GE是AB的垂直平分线,所以GA=GB,同理GD=GC。在△AGD和△BGC中,因为GA=GB,∠AGD=∠BGC,GD=GC,所以△AGD≌△BGC。所以AD=BC。
Reproduce the original title: (in the 2015 middle school mathematics Anhui volume) As shown in Figure 1, in the quadrilateral ABCD, points E and F are the midpoints of AB and CD, respectively, and the point E is the vertical line of AB, and the point F is the CD drape. Lines, two vertical lines intersect at point G, connect AG, BG, CG, DG, and ∠AGD=∠BGC. (1) Verification: AD = BC; (2) Verification: △ AGD ∽ △ EGF; (3) As shown in Fig. 2, if the straight lines AD and BC are perpendicular to each other, find the value of (AD/EF). (1) Proof: Since GE is the vertical bisector of AB, GA=GB and GD=GC. In ΔAGD and ΔBGC, because GA=GB, ∠AGD=∠BGC, and GD=GC, ΔAGD≌ΔBGC. So AD=BC.