ABC is a triangle with D and E as the mid points of the sides AC and AB respectively. G and F are points on side BC such that DG is parallel to EF. Prove that the area of triangle ABC=2xx area of quadrilateral DEFG.?

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2 Answers
May 3, 2018

see explanation.

Explanation:

enter image source here
Let AD=a, => DC=a
let AE=b, => EB=b,
given D and E are midpoint of AC and AB, respectively,
=> DE // CB, and DE=1/2CB
let H be the midpoint of CB,
let CB=2d, => CH=HB=d, => DE=d
draw a line joining D and H, and a line joining E and H, as shown in the figure.
As AD=DC=a, DE=CH=d, and DE // CH,
=> DeltaADE and DeltaDCH are congruent,
=> DH=AE=b,
similarly, DeltaADE and DeltaEHB are congruent,
=> EH=AD=a,
=> DeltaHED, DeltaADE, DeltaDCH and DeltaEHB are all congruent,
let |ABC| denote area of DeltaABC
let |ADE|=x, => |ABC|=4x
=> |CDEH|=|DCH|+|HED|=2x,
given that DG // EF,
=> DeltaDCG and DeltaEHF are congruent
=> |DEFG|=|CDEH|=2x,
=> |ABC|=2xx|DEFG|=2xx2x=4x

Hence, |ABC|=2xx|DEFG|

May 3, 2018

see explanation.

Explanation:

Solution 2:
enter image source here
Given that D and E are the midpoint of AC and AB, respectively, => DE // CB and DE=1/2CB,
=> DeltaADE and DeltaACB are similar,
let |ABC| denote area of DeltaABC,
=> |ADE| : |ACB|= DE^2:CB^2=1:4
let |ADE|=x, => |ACB|=4x, => color(red)(|DEBC|=4x-x=3x)
given DG // EF, => DEFG is a parallelogram,
=> GF=DE=1/2CB=d,
=> |DEBC|=(DE+CB)/2*h=(d+2d)/2*h=(3dh)/2
=> |DEFG|=GF*h=dh
=> |DEFG| : |DEBC|=2:3,
=> |DEFG|=2x,

=> |ABC|=2xx|DEFG|=4x