Question

D, E, F are the mid-point of the sides BC, CA and AB respectively of Δ ABC. Then Δ DEF is congruent to triangle
A. ABC
B. AEF
C. BFD, CDE
D. AFE, BFD, CDE

Answer 1

Given:D, E, F are the mid-point of the sides BC, CA and AB respectively of Δ ABC.

 

To prove : Δ DEF is congruent to triangle

 

Proof:

Since E and F are the midpoints of AC and AB.

BC||FE & FE= ½ BC= BD

(By mid point theorem)

BD || FE & BD= FE

Similarly, BF||DE & BF= DE

Hence, BDEF is a parallelogram

[A pair of opposite sides are equal and parallel]

Similarly, we can prove that FDCE & AFDE are also parallelograms.

Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.

∴ ar(ΔBDF) = ar(ΔDEF) — (i)

In Parallelogram AFDE

ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii)

In Parallelogram FDCE

ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii)

From (i), (ii) and (iii)

ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv)

If area of ∆'s are equal then they are congruent.

Hence , Δ DEF is congruent to triangle ΔBDF = ΔAFE = ΔCDE.

Among the given options option (D) AFE, BFD, CDE is correct.  

HOPE THIS ANSWER WILL HELP YOU……

 

Some more questions :  

If ABC and DEF are two triangles such that Δ ABC≅ Δ FDE and AB =5 cm, ∠B = 40° and ∠A =80°, Then, which of the following is true?

A. DF = 5 cm, ∠F= 60°

B. DE = 5 cm, ∠E= 60°

C. DF = 5 cm, ∠E= 60°

D. DE = 5 cm, ∠D= 40°

https://brainly.in/question/15907780

 

In Δ PQR ≅ Δ EFD then ED =

A. PQ

B. QR

C. PR

D. None of these

https://brainly.in/question/15907770

Answer 2

Given :   D,E,F are the midpoints of the sides BC , CA , AB respectively of triangle ABC  

To Find :  Δ DEF is congruent to triangle

Solution:

Line joining the mid point of two sides of a triangle is half of third side

D is mid point of  side BC  & E is mid point of side CA

hence DE   =  AB/2

E is mid point of side CA & F is mid point of AB

=> EF   =  BC/2

D is mid point of  side BC  &  F is mid point of AB

=> DF =  AC/2

=> DE/AB  =  EF/BC = DF/AC  = 1/2

Two triangles having corresponding sides proportional are similar

=> ΔDEF ≈ Δ ABC

ΔDEF  &  ΔAFE

EF   = EF

DE =  AF  ( DE = AB/2  ,  AF = AB/2)

DF = AE  ( DF = AC/2  , AE = AC/2)

=> ΔDEF  ≅  ΔAFE

Similarly  ΔDEF  ≅  ΔFBD

ΔDEF  ≅  ΔEDC

ΔDEF  ≅  ΔAFE   ≅  ΔFBD ,  ≅  ΔEDC

None of the give option is correct  

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