Q. 25.0( 2 Votes )

D, E and F are the mid-points of, and respectively in ΔABC. Prove that the area of BDEF = 1/2 area of ΔABC.

Answer :

We have


Given: D is the midpoint of , E is the midpoint of & F is the midpoint of .


AF = FB, AE = EC & BD = DC


To Prove: Area of BDEF = 1/2 Area of ∆ABC


Proof: Since, given is that D, E and F are midpoints of respectively.


We know, AE = EC.


AE = FD [, EC = FD]


[, ] …(i)


Also, we know that, AF = FB.


AF = ED [, FB = ED]


[, ] …(ii)


Now, in ∆AFE and ∆DEF, we have


AF ED, AE FD & FE EF


By SSS theorem of congruence, we can say that the congruency AFE DEF is a similarity.


Similarly, in ∆AFE and ∆FBD, we have


AF FB, AE FD & FE BD


By SSS theorem of congruence, we can say that the congruency AFE FBD is a similarity.


Similarly, the correspondence AEF EDC is a similarity.


So, ∆AFE, ∆DEF, ∆FBD and ∆EDC are all congruent triangles.


∆AFE = ∆DEF = ∆FBD = ∆EDC …(iii)


Now,


BDEF = ∆DEF + ∆FBD


BDEF = ∆AFE + ∆AFE [from equation (iii)]


BDEF = 2 ∆AFE …(iv)


And


∆ABC = ∆AFE + ∆DEF + ∆FBD + ∆EDC


∆ABC = ∆AFE + ∆AFE + ∆AFE + ∆AFE [from equation (iii)]


∆ABC = 4 ∆AFE


∆ABC = 2 (2 ∆AFE) …(v)


Comparing equations (iv) and (v), we get


∆ABC = 2 (BDEF)


BDEF = 1/2 ∆ABC


Area of BDEF = 1/2 (Area of ∆ABC)


Thus, proved.


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