Question

In the given figure, $AD\parallel BE\parallel CF$ and $AB=BC$. The ratio of $BE:(AD+CF)=$

• A. 2:1
  
• B. 1:2
  
• C. 3:1
  
• D. 1:1

Answer 1

The correct option is B

1:2

(i)Join DC to meet EB at P
$In△ACD$
(ii) B is mid point of AC (given)
$BP\parallel AD$ (given)
(iii)$\therefore$ P is mid point of DC (Converse of mid point theorem)
(iv) Also $BP=\frac{1}{2}AD$ (mid-point theorem)
In $△DCF$
(v) P is mid-point of DC (from (iii))
(vi) $PE\parallel CF$ (given)
(vii) $\therefore$ E is mid-point of DF(Converse of mid-point theorem)
(viii) $\therefore PE=\frac{1}{2}CF$ (mid-point theorem)
(ix) $BE=BP+PE$
$BE=\frac{1}{2}AD++\frac{1}{2}CF$
$BE=\frac{1}{2}(AD+CF)$
$\therefore \frac{BE}{AD+CF}=\frac{1}{2}$
$BE:(AD+CF)=1:2$
Hence (B)