E and F are respectively the mid points of equal sides AB and AC of ΔABC
SHOW THAT BF = CE
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6
In ∆BCF and ∆CBE
BE=CF (Median )
angle BOE=angle COF (vertically opposite angle
BC=CB (common
So, ∆BCF=~∆CBE
Them, BF=CE(by C. P. C. T)
Hence proved
BE=CF (Median )
angle BOE=angle COF (vertically opposite angle
BC=CB (common
So, ∆BCF=~∆CBE
Them, BF=CE(by C. P. C. T)
Hence proved
syedkamran2913:
thanks mate i got it
bro btw in triangle bcf or cbe angle boe dosent exist
Its axist if you will form
it can bro but even if u form u cant get bc common
Answered by
2
In ∆ ABF and ∆ ACE
AB=AC
<A=<A
AF=AE
∆ABF~=∆ACE
BF=CE
AB=AC
<A=<A
AF=AE
∆ABF~=∆ACE
BF=CE
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