prove that BE =CF
pls pls
Answers
Step-by-step explanation:
it is easiest method, i hope this will help you
please mark me as brainliest it's a humble request.
In the Δs ABE, ACF
AB=AC given
AE=AF half of AC, AB resp.
∠BAE=∠CAF common
Therefore ΔABE≅ΔACF two sides and included angle
since both triangles are congruent, then there all sides will be of same length
In particular, BE=CF
You don't need to prove FE∥BC, nor to quote the base angles of an isoceles triangle theorem.
In fact the same proof applies whenever points E and F are equidistant from A on the sides AC,AB (not necessarily the mid points). This is used in Euclid's proof of Proposition 5 (the base angles of an isoceles triangle theorem), though he could, more elegantly, have used the fact that ΔABC≅ΔACB (two sides and the included angle).