# 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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  RD Sharma | Understanding Shapes (Polygons) Part 142 mins  Basic Understanding of Polygons44 mins  NCERT | Surface Area of Solid Shapes39 mins  Quadrilaterals32 mins  Quiz | Visualising Solid Shapes36 mins  Genius Quiz | Understanding Quadrilaterals20 mins  Quiz | Polygon & their Properties46 mins  NCERT | Understanding Important Laws49 mins  Champ Quiz | Cell44 mins  Understanding Quadrilateral43 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 