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

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.