Question

1 如图101所示,矩形ABCD的对角线相交于点O,BE⊥DE于点E,求证AE⊥CE

Answer 1

连接OE

BE⊥DE,BED是直角三角形

OE = 1/2BD

矩形对角线相等,且相互平分

OB=OD=OA=OC

所以 OE = OA =OC

EOA,EOC是等腰三角形

∠OEA+∠OEC= 1/2(∠OAE+∠OEA+∠OEC+∠OCE) = 1/2*180° = 90°

2

BF是角平分线

FBC =45°.

BFC= 45°

所以 BC=FC

OBC = 45+15 =60°

OB=OC,所以 OBC是等边三角形,BC=OC

所以 OC=FC

∠CFO = ∠COF

∠OCF = 90° -60° =30°

∠COF = 1/2 (180°-30°) = 75°

∠OEF = ∠ECF+ ∠EFC = 30°+45° = 75°

所以 OF =EF