prove AF=1\3AC plzz with exp
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draw DG which is parallel with BF
in ∆BCF,D is the mid point of BC
BF PARALLEL TO DG
G IS THE MID POINT OF FC
FG=GC. IN ADG,F IS THE MID POINT OF AG. AF=FG
AF=FG=GC
AC=AF+FG+GC=3AF
AF=1/3AC
in ∆BCF,D is the mid point of BC
BF PARALLEL TO DG
G IS THE MID POINT OF FC
FG=GC. IN ADG,F IS THE MID POINT OF AG. AF=FG
AF=FG=GC
AC=AF+FG+GC=3AF
AF=1/3AC
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