E and F are representative the midpoint of equal sides ab and ac of triangle ABC show that BF equal to see
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anishbaxi1787:
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hi there
in triangle ebc and triangle fcb
eb = fc (1÷2ab and 1÷2 ac)
angle ebc=fcb. (able opposite to daal sides)
bc= cb. (common)
so. triangle ebc and triangle fcb are congurent by SAS
so bf= ce (by CPCT)
in triangle ebc and triangle fcb
eb = fc (1÷2ab and 1÷2 ac)
angle ebc=fcb. (able opposite to daal sides)
bc= cb. (common)
so. triangle ebc and triangle fcb are congurent by SAS
so bf= ce (by CPCT)
click on thank you
And mark in brainlist
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