Question : $B_{1}$ is a point on the side AC of $\Delta ABC$ and $B_{1}B$ is joined. A line is drawn through A parallel to $B_{1}B$ meeting BC at $A_{1}$ and another line is drawn through C parallel to $B_{1}B$ meeting AB produced at $C_{1}$. Then:
Option 1: $\frac{1}{CC_{1}}-\frac{1}{AA_{1}}=\frac{1}{BB_{1}}$
Option 2: $\frac{1}{CC_{1}}+\frac{1}{AA_{1}}=\frac{1}{BB_{1}}$
Option 3: $\frac{1}{BB_{1}}-\frac{1}{AA_{1}}=\frac{2}{CC_1}$
Option 4: $\frac{1}{AA_1}-\frac{1}{CC_1}=\frac{2}{BB_1}$
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Correct Answer: $\frac{1}{CC_{1}}+\frac{1}{AA_{1}}=\frac{1}{BB_{1}}$
Solution :
In $\triangle AA_{1}C$ and $\triangle BB_{1}C$
$BB_1 \left | \right |AA_1$
$\frac{BB_1}{AA_1}=\frac{B_1C}{AC}$............(equation 1)
In $\triangle ACC_{1}$ and $\triangle ABB_{1}$
$BB_1 \left | \right |CC_1$
⇒ $\frac{BB_1}{CC_1}=\frac{AB_1}{AC}$
⇒ $\frac{BB_1}{CC_1}=\frac{AC-B_1 C}{AC}$
⇒ $\frac{BB_1}{CC_1}=1-\frac{B_1 C}{AC}$
Putting the value from equation 1, we get:
$\frac{BB_1}{CC_1}=1-\frac{BB_1}{AA_1}$
⇒ $BB_1[\frac{1}{CC_1}+\frac{1}{AA_1}]=1$
⇒ $\frac{1}{CC_{1}}+\frac{1}{AA_{1}}=\frac{1}{BB_{1}}$
Hence, the correct answer is $\frac{1}{CC_{1}}+\frac{1}{AA_{1}}=\frac{1}{BB_{1}}$.
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