DEF are mid point of the sides BC,CAandABrespectively of TRIANGLE ABCthen triangle DEF is congruent to triangle
Answers
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.
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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°
brainly.in/question/15907780
In Δ PQR ≅ Δ EFD then ED =
A. PQ
B. QR
C. PR
D. None of these
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