❣️Brainly Challenge❣️
27. In a trapezium ABCD , AB|| DC and DC=2AB. EF||AB , where E and F lie on BC and AD respectively such that BE/EC=4/3. Diagonal DB intersect EF at G .Prove that ,
7EF=11AB.
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Step-by-step explanation:
From figure:
(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
⇒ 7 EF = 11 AB.
Hope it helps!
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