Question

In trapezium ABCD, AB||CD, DC=2AB. EF||AB cuts AD in F and BC in E. {BE}/{EC}={4}/{3} . DB intersects EF at G .PROVE THAT 7EF=11AB

Answer 1

Answer:

Step-by-step explanation:

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!