In Fig. 17.29, suppose it is known that DE = DF.

In $\Delta$ FDE :

DE $=$ DF

$\therefore \angle \mathrm{FED}=\angle \mathrm{DFE} \ldots \ldots \ldots \ldots$ (i) (angles opposite to equal sides)

In the IIg BDEF :

$\angle \mathrm{FBD}=\angle \mathrm{FED} \ldots \ldots$ (ii) (opposite angles of a parallelogram are equal)

In the II $^{\mathrm{gm}}$ DCEF:

$\angle \mathrm{DCE}=\angle \mathrm{DFE} \ldots$... (iii) (opposite angles of a parallelogram are equal)

From equations (i), (ii) and (iii):

$\angle \mathrm{FBD}=\angle \mathrm{DCE}$

In $\triangle \mathrm{ABC}:$

If $\angle \mathrm{FBD}=\angle \mathrm{DCE}$, then $\mathrm{AB}=\mathrm{AC}$ (sides opposite to equal angles).

Hence, $\triangle \mathrm{ABC}$ is isosceles.