In the given figure DE,EF,DF are line segments drawn parallel to AB,BC,CA respectively of triangle ABC and through its vertices
Show that
b) A is mid-point of FE
(ii) perimeter of A ABC
1/2(perimeter of A DEF)
Answers
(1) Since E and F are the midpoints of AC and AB.
BC||FE & FE= ½ BC= BD
BD || FE & BD= FE ( By mid point theorem)
Similarly,
BF||DE & BF= DE
Hence, BDEF is a parallelogram
(2) Similarly, we can prove that FDCE & AFDE are also parallelogram
Since,
BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.
∴ ar(ΔBDF) = ar(ΔDEF) ......... (1)
In Parallelogram AFDE
ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) ......... (2)
In Parallelogram FDCE
ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) ......... (3)
From (1), (2) and (3)
ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF) .....(4)
ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)
4 ar(ΔDEF) = ar(ΔABC) (From eq 4)
ar(∆DEF) = 1/4 ar(∆ABC) ........(5)
(3) Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)
ar(parallelogram BDEF) = 2× ar(ΔDEF) (From eq 4)
ar(parallelogram BDEF) = 2× 1/4
ar(ΔABC) (From eq 5)
ar(parallelogram BDEF) = 1/2 ar(ΔABC)
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