Question

in a Trapezium ABCD AB parallel to CD. DC equal twice of AB FE parallel to AB cuts AD in F and BC in E such that BE/EC is equal 3/4.diagonal BD intersect EF at G. prove that 7FE=10AB.

Answer 1

In trapezium ABCD, AB||DA and DC=2AB. FE drawn parallel to AB cuts AD in F  and BC in E such that  BE

/EC  =  3/4. Diagonal DB intersects EF at G.

Prove that 7FE = 10 AB.

Explanation:

In △DFG and △DAB, we have

   ∠1=∠2     [ ∵ AB ∣∣ DC∣∣ EF ∴ ∠1 and ∠2 are corresponding angles]

   ∠ FDG = ∠ ADB   [Common]

So, by AA - criterion of similarity,

we have

∴  Δ DFG∼Δ DAB

⇒  DADF​=ABFG​.......(i)

In trapezium ABCD, we have

    EF||AB||DC

∴   DFAF​=ECBE​

⇒  DFAF​=43​

[∵ECBE​=43​(Given)

⇒ DFAF​+1=43​+1      [Adding 1 on both sides]

⇒  DFAF+DF​=47​

⇒  DFAD​=47​

⇒  ADDF​=74​......(ii)

From (i) and (ii), we get

  ABFG​=74​

⇒ FG=74​AB........(iii)

In Δ BEG and Δ BCD, we have

   ∠BEG=∠BCD        [Corresponding angles]

   ∠B=∠B             [Common]

∴  Δ BEG ∼ Δ BCD      [By AA-criterion of similarity]

⇒  BCBE​ = CDEG​

⇒  73 ​= CDEG​

[ ∵ ECBE​ = 43 ​⇒  BEEC​ = 34​ ⇒ BEEC​ +1 = 34​+1 ⇒ BEBC​ = 37 ​]

⇒  EG = 73​CD

⇒  EG = 73 ​× 2AB

⇒ EG = 76​ × AB

Adding (iii) and (iv), we get

 FG + EG = 74​AB + 76​AB

⇒ EF = 710​AB

⇒ 7EF = 10AB  ⇒   Hence proved