Question

Eand F are respectively the mid point of the side AB and AC of∆ABC show that BF=CE

Answer 1

(if ABC is an isosceles∆ )

in ∆BEC & ∆CFB

BC=BC. ( base is common)

EB=FC. (isosceles∆)

angle E= angle F ( angles between the same parallels)

So, both the ∆s are congruent by SAS congruency rule,

BF=CE. ( cpct)