Question

In a trapezium ABCD, AB//CD and DC= 2AB. EF//AB where E AND F lie on BC and AD RESPECTIVELY. BE/EC= 3/4 WHERE DIAGONAL DB INTERSECTS EF AT G . Prove that 7EF = 11 AB.

Answer 1

There is a problem in the given exercise.  It should be 7 EF = 10 AB.  If it is given that BE / EC = 4/3, then  it would be 7 EF = 11 AB.

See the diagram. 

Draw BH || AD.  So ABHD and ABIF are parallelograms.  DH = AB  = FI
As,  DC = 2 AB ,   DH = HC = AB

ΔBEI and ΔBCH are similar, as corresponding sides are parallel. BI || BH, BE || BC and IE || HC.

IE / HC = BE / BC = 3 / (3+4) = 3/7

IE = 3 HC / 7

EF = FI + IE = DH + 3 HC / 7
     = AB + 3 AB/7
     = 10/7 * AB

7 EF = 10 AB

Answer 2

$<b>Step-by-step explanation:</b>$

$<u>From figure:</u>$

(i)

In ΔDFG and ΔDAB, we have

⇒ ∠FDG = ∠ADB

∴ ΔDFG ~ ΔDAB

⇒ (DF/DA) = (FG/AB)

(ii)

In trapezium ABCD, EF ║ AB ║ DC

⇒ (AF/DF) = (BE/EC)

⇒ (AF/DF) = 4/3

⇒ (AF/DF) + 1 = 4/3 + 1

⇒ (AF + DF)/DF = 7/3

⇒ AD/DF = 7/3

⇒ DF/AD = 3/7  

(iii)

From (i) & (ii), we have

⇒ FG/AB = 3/7

⇒ FG = (3/7) AB

(iv)

From ΔBEG and ΔBCD, we have

⇒ ∠BEG = ∠BCD

∴ ΔBEG ~ ΔBCD

⇒ (BE/BC) = EG/CD

⇒ 4/7 = EG/CD

⇒ EG = (4/7) CD

⇒ EG = (4/7) * 2AB

⇒ EG = (8/7) AB

On solving (iii) & (iv), we get

⇒ FG + EG = (3/7) AB + (8/7) AB

⇒ EF = (11/7)AB

⇒ $\sf{\bold{\large{7\: EF = 11\: AB.}}}$

Hope it helps!