第1个回答 2016-07-23
已知:AB∥CD∥EF,GI,JL交AB,CD,EF于点G,J,H,K,I,L.求证:GH:HI=JK:KL证明:过点K作G'I'∥GI交AB,CD,EF于点G',H'I'.∵AB∥CD∥EF,G'I'∥GI∴ 四边形GHKG',HII'H‘,GII'G是平行四边形(平行四边形判定定理),∠BJK=∠KLI,∠JG'I'=∠G'I'F(内错角相等)∴△JG'K∽△I'LK,(相似三角形判定),GH=G'H',HI=H'I'(平行四边形对边相等)∵G'H':H'I'=JK:KL(相似三角形性质)∴GH:HI=JK:KL(等量代换)