Congruency of Triangles, To prove BE = CF

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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)