如图,△ABC中,AD是中线,AE是角平分线,CF⊥AE于F,AB=5,AC=2,则AFEF=
<p>解:延长CF交AB于点G,<br />∵AE平分∠BAC,<br />∴∠GAF=∠CAF,<br />∵AF垂直CG,<br />∴∠AFG=∠AFC,<br />在△AFG和△AFC中,<br />∵</p><p> ∠GAF=∠CAFAF=AF∠AFG=∠AFC </p>,<br />∴△AFG≌△AFC(ASA),<br />∴AC=AG,GF=CF,<br />又∵点D是BC中点,<br />∴DF是△CBG的中位线,<br />∴DF=1/2<p> <br /></p>BG=1/2<p><br /></p>(AB-AG)=1/2<p><br /></p>(AB-AC)=3/2<p><br /></p>故答案为:3/2<p><br/></p><p><br /></p>
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