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					 B1 is a point on the side AC of ∆ABC and B1B is joined. A line is drawn through A parallel to B1B meeting BC at A1 and another line is drawn through C parallel to B1B meeting AB produced at C1. Then
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                        -  1 - 1 = 1 CC1 AA1 BB1 
-  1 + 1 = 1 CC1 AA1 BB1 
-  1 + 1 = 2 BB1 AA1 CC1 
-  1 + 1 = 2 AA1 CC1 BB1 
 
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Correct Option: B
According to question , we draw a figure
In ∆ AA1C and ∆BB1C, 
BB1 || AA1 ⇒ ∆AA1C ~ ∆BB1C
| ∴ | = | ..... (i) | ||
| BB1 | B1C | 
In ∆ ACC1 and ∆ ABB1,
BB1 || CC1 ⇒ ∆ACC1 ~ ∆ABB1
| ∴ | = | ||
| BB1 | AB1 | 
| ⇒ | = | = | |||
| CC1 | AC | AC | 
| ⇒ | = 1 - | ||
| CC1 | AC | 
| ⇒ | = 1 - | [From equation (i) | ||
| CC1 | AA1 | 
| ⇒ | + | = 1 | ||
| CC1 | AA1 | 
| ⇒ | + | = | |||
| CC1 | AA1 | BB1 | 
 
	