如图,△ABC为等腰三角形,∠ACB=90°,延长AB至F,使∠ECF=135°.求证:AE:EC=BA:CF.

问题描述:

如图,△ABC为等腰三角形,∠ACB=90°,延长AB至F,使∠ECF=135°.求证:AE:EC=BA:CF.

AE:EC=BA:CF.BA?ABEF(共线)△ABC为等腰三角形,∠ACB=90° 得BC=AC ∠CBA=45° ∠CAB=45° ∠EAC=135° ∠CBF=135°∠EAC=135° ∠AEC+ ∠ECA=180°-135° =45° ∠CBF=135° ∠BCF+ ∠CFB=180°-135° =45°∠ECF=1...