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In Figure, suppose it is known that DE = DF. Then, is ΔABC isosceles? Why or why not? |
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Answer» In parallelogram BDEF BD = EF and BF = DE [opposite sides are equal in a parallelogram] In parallelogram DCEF DC = EF and DF = CE [opposite sides are equal in a parallelogram] In parallelogram AFDE AF = DE and DF = AE [opposite sides are equal in a parallelogram] So, DE = AF = BF Similarly: DF = CE = AE Given, DE = DF Since, DF = DF AF + BF = CE + AE AB = AC ∴ ΔABC is an isosceles triangle. |
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