Question

pls help to solve this problem by using mid-point theorem

prove BE = 1/2AC​

Answer 1

Answer:

according to the midpoint theorem he is equal to C in the bec EF divide bc into two equal halves that is BF and CF

Step-by-step explanation:

In triangle bef and triangle efc

bf=cf (by mid point theorem)

ef=ef (common)

angle bef =angel cef (angle opposite to equal sides)

bef is congruent to efc (sas)

be=ec (CPCT) .......1

AE= Ec......... 2

from 1 and 2

be =ae, hence proved

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Answer 2

Step-by-step explanation:

according to the midpoint theorem he is equal to C in the bec EF divide bc into two equal halves that is BF and CF

Step-by-step explanation:

In triangle bef and triangle efc

bf=cf (by mid point theorem)

ef=ef (common)

angle bef =angel cef (angle opposite to equal sides)

bef is congruent to efc (sas)

be=ec (CPCT) .......1

AE= Ec......... 2

from 1 and 2

be =ae, hence proved

please mark as brainliest