Question

In a trapezium ABCD , AB parallel to CD and DC=4AB. Also EF parallel to AB intersect DA and CB at E and F such that AE÷ED =3/4 . Prove that 5EF = 14 AB

Answer 1

There seems to be some error in the given exercise.  Is that really 5 EF = 14 AB?

See the diagram.    We use similar Δs.

ΔAJD and  ΔAGE are similar.   AG / AJ = GE / JD   --(1)

ΔBHF and ΔBIC are similar.
     HF / IC = BH / BI = BF / BC  = 3 : (4+3) = 3/7   --(2)

Quadrilaterals AGHB and AJIB are rectangles as the corresponding sides  are || and internal angles are 90°. 
So AG / AJ = BH/BI
Hence from (1) and (2) we get:

GE / JD = HF / IC = 3 / 7
    or,   JD =  7 GE /3   &   IC = 7 HF /3

Now   DC = DJ + JI + IC
=>   4 AB  = AB + 7 (HF  + GE) /3
=>   9/7 * AB = GE + HF
=>   EF = GH + GE + HF
            = AB + 9 AB/7 = 16 AB / 7 
=>   EF = 4  DC/7

=>   7 EF = 4 DC = 16 AB.

Answer 2

Answer:

Step-by-step explanation:

There seems to be some error in the given exercise.  Is that really 5 EF = 14 AB?

See the diagram.    We use similar Δs.

ΔAJD and  ΔAGE are similar.   AG / AJ = GE / JD   --(1)

ΔBHF and ΔBIC are similar.

     HF / IC = BH / BI = BF / BC  = 3 : (4+3) = 3/7   --(2)

Quadrilaterals AGHB and AJIB are rectangles as the corresponding sides  are || and internal angles are 90°. 

So AG / AJ = BH/BI

Hence from (1) and (2) we get:

GE / JD = HF / IC = 3 / 7

    or,   JD =  7 GE /3   &   IC = 7 HF /3

Now   DC = DJ + JI + IC

=>   4 AB  = AB + 7 (HF  + GE) /3

=>   9/7 * AB = GE + HF

=>   EF = GH + GE + HF

            = AB + 9 AB/7 = 16 AB / 7 

=>   EF = 4  DC/7

=>   7 EF = 4 DC = 16 AB.

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