# Medians BE and CF of a triangle ABC are equal. Prove that the triangle is isosceles

Given
Medians BE  = CF , of a $∆$ ABC ,
Let BE and CF meet at G the centroid .

In $∆$ ECG And $∆$ ​FBG

GE = $\frac{BE}{3}$
And
GF = $\frac{CF}{3}$    ( As we know medians are trisected at centroid  )
SO,

GE = $\frac{BE}{3}$  = GF = $\frac{CF}{3}$                    ( Given BE  = CF )

And
GC = $\frac{2CF}{3}$  = GB  = $\frac{2BE}{3}$    ( As we know medians are trisected at centroid and BE  = CF is given )
And
$\angle$ EGC   =  $\angle$ FGB                    ( opposite angles )
Hence
$∆$ ECG $\cong$$∆$ ​FBG                      ( By SAS rule )
So,
BF = CE                                     ( CPCT )
But BF  = $\frac{AB}{2}$  And  CE  = $\frac{AC}{2}$  ( As we know BE and CF are medians , So E and F are  the midpoints of AB and AC )
Hence
AB  = 2 BF = AC = 2CE
SO,
AB  = AC
Hence
$∆$ ABC is a isosceles triangle                                     ( Hence proved )

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