$E$ and $F$ are respectively the mid-points of equal sides $AB$ and $AC$ of $\Delta ABC$ (see figure). Show that $BF = CE$.
(N/A) In $\Delta ABF$ and $\Delta ACE$:
$AB = AC$ (Given)
$\angle A = \angle A$ (Common angle)
$AF = AE$ (Since $F$ and $E$ are mid-points of equal sides $AC$ and $AB$ respectively,$AF = \frac{1}{2} AC$ and $AE = \frac{1}{2} AB$. Since $AB = AC$,it follows that $AF = AE$.)
Therefore,$\Delta ABF \cong \Delta ACE$ by the $SAS$ congruence rule.
Since the triangles are congruent,their corresponding parts are equal.
Thus,$BF = CE$ (by $CPCT$).
$AB = AC$ (Given)
$\angle A = \angle A$ (Common angle)
$AF = AE$ (Since $F$ and $E$ are mid-points of equal sides $AC$ and $AB$ respectively,$AF = \frac{1}{2} AC$ and $AE = \frac{1}{2} AB$. Since $AB = AC$,it follows that $AF = AE$.)
Therefore,$\Delta ABF \cong \Delta ACE$ by the $SAS$ congruence rule.
Since the triangles are congruent,their corresponding parts are equal.
Thus,$BF = CE$ (by $CPCT$).
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