In Fig. 17.29, suppose it is known that DE = DF. Then, is ΔABC isosceles? Why or why not?

In parallelogram BDEF

BD = EF and BF = DE ……..(i) [In a parallelogram opposite sides are equal]

In parallelogram DCEF

DC = EF and DF = CE ……..(ii) [In a parallelogram opposite sides are equal]

In parallelogram AFDE

AF = DE and DF = AE ……..(ii) [In a parallelogram opposite sides are equal]

Therefore DE = AF =BF ……..(iv)

Similarly: DF = CE = AE …….(v)

But, DE = DF ……given

From equations (iv) and (v), we get

AF + BF = AE + EC

AB = AC

Therefore ΔABC is an isosceles triangle.