Congruency of Triangles, To prove BE = CF I am preparing for my maths exam on saturday, I could not find a satisfactory answer to a question I am stuck on.
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Given an isosceles triangle ABC in which AB = AC. If E and F be midpoints of AC and AB respectively, prove BE = CF.
I know the constructed line EF is parallel to BC but I could not prove the same satisfactorily. Thanks.
 A: The base angles of an isosceles triangle are equal. SAS proves triangles are equal. Then corresponding parts of congruent triangles are equal finishes the proof.
A: The proof is like this (by congruency):
In $\triangle BEC$ and $ \triangle CFB$
$$BC \cong CB $$
$$\angle ABC \cong\angle ECB$$
$$AB\cong AC \Leftrightarrow BF \cong EC$$
By $SAS$ congruence criterion
$$\triangle BEC \cong \triangle CFB$$
So $BE \cong CF $ (Corresponding parts of Congruent Triangles)
A: In the $\Delta$s ABE, ACF
$AB=AC$ given
$AE=AF$ half of AC, AB resp.
$\angle BAE=\angle CAF$ common
Therefore $\Delta ABE\cong \Delta ACF$ two sides and included angle
In particular, $BE=CF$
$QED$
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 $\Delta ABC\cong \Delta ACB$ (two sides and the included angle). 
