Question

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

Answer 1

\[\text{ In } ∆ FDE: \]
\[DE = DF \]
\[ \therefore \angle FED = \angle DFE . . . . . . . . . . . . . (i) (\text{ angles opposite to equal sides })\]
\[\text{ In the } {II}^{gm} BDEF: \]
\[\angle FBD = \angle FED . . . . . . . (ii) (\text{ opposite angles of a parallelogram are equal })\]
\[\text{ In the } {II}^{gm} DCEF: \]
\[\angle DCE = \angle DFE . . . . . . (iii) (\text{ opposite angles of a parallelogram are equal })\]
\[\text{ From equations } (i), (ii) \text{ and } (iii): \]
\[\angle FBD = \angle DCE\]
\[\text{ In } \bigtriangleup ABC: \]
\[If \angle FBD = \angle DCE, \text{ then } AB = AC (\text{ sides opposite to equal angles }) . \]
\[\text{ Hence }, \bigtriangleup ABC \text{ is isosceles }.\]