Question

矩形ABCD.点O为AC中点.AC=2AB,延长AB至G使BG=AB.连接GO交BC于E延长GO交AD于F证明四边形AE

Answer 1

证明：连接CG,∵在矩形ABCD中AC=2AB,BG=AB,∴AG=AC,∠CAG=60?∴△ACG是等边三角形,∵O为AC的中点,∴GF⊥AC,∵在矩形ABCD中,BC‖AD,∴∠DAC=∠BCA,AO=OC,∠AOF=∠COE=90?∴△AOF≌△COE,∴CE=AF,∴四边形AECF是平行四边形,∴四边形AECF是菱形（对角线互相垂直的平行四边形是菱形）．