Question

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(1)  ar ( ADEG) = ar (GBCD)

 (2)  ar (ΔEGB) = `1/6` ar (ABCD)

 (3)  ar (ΔEFC) = `1/2` ar (ΔEBF)

 (4)  ar (ΔEBG)  = ar (ΔEFC)

 ⁽⁵⁾ΔFind what portion of the area of parallelogram is the area of EFG.

Answer 1

Given,
ABCD is a parallelogram

AG = 2GB,CE = 2DE and BF = 2FC
To prove :

(1) ar (ADEG) = ar (GBCE)

(2) ar (Δ EGB) = `1/6` are (ABCD)

(3) ar (Δ EFC) =`1/2` area  (Δ EBF)

(4) area  (ΔEBG) = 3/2 area  (EFC)

(5) Find what portion of the area of parallelogram is the area of ΔEFG.

Construction: draw EP ⊥ AB and  EQ ⊥ BC

Proof : we have,

  AG = 2GB and CE = 2DE and BF = 2FC 

⇒ AB - GB = 2GB and CD - DE =  2DE and BC - FC  

⇒ AB -GB = 2GB and CD - DE = 2DE and BC - FC = 2FC.

⇒ AB =3GB  and CD =3DE and BC = 3FC

⇒ `GB = 1/3 AB and DE = `1/3` CD and FC =  1/3 BC `   ............ (1)

(1)  ar (ADEG) `1/2` (AG + DE) × EP

⇒  ` ar (ADEG) = 1/2 (2/3AB + 1/3CD) xx EP` [By                  using(1)]

⇒  `ar (ADEG) = 1/2 (2/3 AB + 1/3 AB)xx EP`    [∴ AB          = CD] 

⇒  `ar (ADEG) = 1/2 xx AB xx EP`      ........... (2)

`And  ar  (GBCE) = 1/2 (GB + CE) xx EP`  

⇒ `ar (GBCE) = 1/2 [1/3 AB + 2/3 CD ] xx EP`  [By using (1)]

⇒  `ar (GBCE) = 1/2 [1/3 AB + 2/3 AB ] xx EP`  [∴ AB = CD]

⇒ `ar (GBCE) = 1/2 xx AB xx EP`    ........ (1) 

Compare equation (2) and (3) 

(2)   ar (ΔEGB) = `1/2xx GB xx EP`

       = `1/6 xx AB xx EB`

        = `1/6 ar (1^(9m) ABCD)`

(3)  `Area (ΔEFC) = 1/2 xx FC xx EQ ......... (4)`

      `And area (ΔEBF) = 1/2 xx BF xx EQ` 

      ⇒ `ar (ΔEBF) = 1/2  xx2 FC xx EQ `        [BF = 2FC given]

      ⇒ `ar (ΔEBF) = FC xx EQ`   ............. (5)

        Compare equation 4 and 5

         `Area (ΔEFC) = 1/2 xx area (ΔEBF) `

 (4)  From (1) part 

      `ar (ΔEGB ) = 1/6 ar (11^(5m )ABCD)` ....... (6) 

          Form (3) part 

       ` ar (ΔEFC) =1/2 ar  (EBF)`

⇒   `ar (ΔEFC)  = 1/3 ar(ΔEBC)`   

⇒ `ar (ΔEFC) = 1/3 xx 1/2 xx CE xx EP` 

`= 1/2 xx 1/3 xx 2/3 CD xx EP`

 `=1/6 xx 2/3 xx ar (11^(gm)ABCD)`

 ⇒ `ar (ΔEFC) = 2/3 xx ar (ΔEGB)`    [By using]

  ⇒ `ar (ΔEGB) = 3/2 ar (ΔEFC). `

 (5)  Area (ΔEFG) = ar (Trap . BGEC) = - ar (ΔBGF)  → (1)

    `Now , area  (trap BGEC)  = 1/2 (GB + EC) xx EP`

 = `1/2 (1/3 AB + 2/3 CD) xx EP `

 = `1/2 AB xx EP`

 = `1/2 ar (11^(5m)ABCD)`  

`Area (ΔEFC) = 1/9  area  (11^(5m)ABCD)`   [Form 4 part]

And area (Δ BGF) = `1/2 BF xx GR`

   `= 1/2 xx 2/3 BC xx GR`

    `= 2/3 xx 1/2 BC xx GR `

    `= 2/3 xx ar (Δ GBC) `

    `= 2/3 xx 1/2 GB xx EP`

     `= 1/3 xx 1/3 AB xx EP`

     `= 1/9 AB xx EP`

     ` 1/9 ar ( 11^(gm) ABCD)`    [ From (1)]     

 `ar ( ΔEFG) = 1/2 ar ( 11^(gm) ABCD) = 1/9 ar ( 11^(gm) ABCD) = 1/9 ar ( 11^(gm) ABCD)`

 `= 5/18 ar  ( 11^(gm) ABCD).`