D e f are respectively the midpoint of side ab BC and CA of triangle ABC find the ratio of the area of triangle d e f and triangle ABC..
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In ∆ABC, D and F are the points of sides AB and CA repectively.
=>DF||BE [by midpoint theorem]
=>DF||BE.
Similarly,EF||BD.
=>BEFD is a parallelogram.
=>angle.B=angle.EF=1/2AB
and DF=BE=1/2BC.
Also,ECFD is a parallelogram.
=>angle.EDF=angle.C
Now,in ∆DEF and ∆CAB,we have
angle.EFD=angle.B
and angle.EDF=angle.C
=>∆DEF~∆CAB [by AA similarity]
And so,ar(∆DEF)/ar(∆ABC) = ar(∆DEF)/ar(∆CAB) = DF^2/BC^2 = (1/2BC)^2/BC^2 = 1/4
=>DF||BE [by midpoint theorem]
=>DF||BE.
Similarly,EF||BD.
=>BEFD is a parallelogram.
=>angle.B=angle.EF=1/2AB
and DF=BE=1/2BC.
Also,ECFD is a parallelogram.
=>angle.EDF=angle.C
Now,in ∆DEF and ∆CAB,we have
angle.EFD=angle.B
and angle.EDF=angle.C
=>∆DEF~∆CAB [by AA similarity]
And so,ar(∆DEF)/ar(∆ABC) = ar(∆DEF)/ar(∆CAB) = DF^2/BC^2 = (1/2BC)^2/BC^2 = 1/4
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