Question

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.?

Answer 1

see explanation.

Explanation:

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|`

Answer 2

see explanation.

Explanation:

Solution 2:

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`