Question

In the given figure, ABC is a triangle with D as the mid point of AB, DE is parallel to AC, DF is parallel to BC. Also DE = DF. Prove that Triangle ABC is an isosceles triangle. ​

Answer 1

Answer:

given:.

in ∆ abc

DC bisects AB and /_C

DE||AC,DF||BC

DE=DF

to prove:

∆abc is isosceles i,e.CA=BC

proof:

in ∆ DCE and ∆DCF

/_DCF=/_DCE. [given]

CD=CD. [commom]

/_CED=/_CFD. [as shown in figure]

therefore,∆DCE=~∆DCF. [ASA------------(i)

now,in ∆DEA and ∆DFB

DA=DB. [given]

/_DEA=/_DFB. [as shown in the figure]

DE=DF. [given]

therefore, ∆DEA=~∆DFB. [SAS]--------------(ii)

now from equation (i) and (ii)

∆DCE+∆DEA =~ ∆DCF+∆DFB

= ∆DCA=~∆DCB

therefore, CA=CB. [by CPCT]

so, ∆abc is isosceles...

hope it helps you