Q. 243.7( 3 Votes )
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.
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