Q. 243.7( 3 Votes )

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


Answer :

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.


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