Q. 25.0( 2 Votes )

D, E and F are th

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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