如图所示.在△ABC中,∠BAC=120°,AD平分∠BAC交BC于D.求证:1/AD=1/AB+1/AC.
问题描述:
如图所示.在△ABC中,∠BAC=120°,AD平分∠BAC交BC于D.求证:
=1 AD
+1 AB
. 1 AC
答
证明:过D引DE∥AB,交AC于E.∵AD是∠BAC的平分线,∠BAC=120°,∴∠BAD=∠CAD=60°.又∠BAD=∠EDA=60°,所以∴△ADE是正三角形,∴EA=ED=AD.①由于DE∥AB,所以△CED∽△CAB,∴DEAB=CECA=CA−AECA=1-AECA.②...