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)
BDEF = ∆DEF + ∆FBD
⇒ BDEF = ∆AFE + ∆AFE [from equation (iii)]
⇒ BDEF = 2 ∆AFE …(iv)
∆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)
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