[" If "Delta DEF~=Delta BCA" ,write the part(s) "Af" ABC that correspond to "],[[" (i) "/_E," (ii) "(2)/(EF)," (iii) "/_F]]

If Delta DEF[-=sim?]Delta BCA, write the part (s) of Delta BCA that correspond to (i) /_E( ii) /_EF (iii)

If Delta ABC ~= Delta DEF write the parts of Delta ABC that correspond to angle F

If DeltaDEF [-=~?] DeltaBCA, write the part (s) of DeltaBCA that correspond to (i) angleE (ii) angle overline(EF) (iii) < F (iv) overline(DF)

If Delta ABC~Delta DEF such that 2AB=DE and BC=6 cm find EF.

If Delta ABC ~ Delta DEF and AB = DE, BC = 8 cm, then the value of EF ?

In Delta ABC and Delta DEF , if AB = EF, BC= DE and CA = FD, then

In Delta ABC , DE || BC and CD || EF . Prove that AD^(2) = AF xx AB

If Delta ABC-Delta DEF,BC=3cm,EF=4cm and area of Delta ABC=54sqcm then find the area of Delta DEF

If Delta ABC ~ Delta DEF such that area of Delta ABC " is " 9 cm^(2) and the area of Delta DEF " is " 16 cm^(2) and BC=2.1 cm. Find the length of EF.

Delta ABC=Delta DEF, Prove that (AB)/(DE)=(AB+BC+AC)/(DE+EF+DF)