Consider the parallelogram \( A B C D \) as shown in the figure, where

1.	Consider the parallelogram \( A B C D \) as shown in the figure, where \( \frac{A E}{A B}=\frac{C F}{C D}=\frac{1}{n} \), for some positive integer \( n \). Suppose the length of \( A C \) is \( a \), then the length of \( X Y \) is A. \( \frac{a}{n} \). B. \( \frac{n a}{n+1} \). C. \( \frac{(n-1) a}{n+1} \). D. \( \frac{(n-1) a}{n} \).
Answer» Correct option is (B) \( \frac{na}{n + 1}\) ∵ ABCD is a parallelogram ∴ AB = CD ∴ \(AE = \frac{AB}n = \frac{CD}n = CF\) \(\left(\because \frac{AE}{AB} = \frac{CF}{CD} = \frac 1n (given)\right)\) In \(\triangle CDX\), \(\triangle CFY \sim\triangle CDX\) \(\therefore FY \parallel DX\) ⇒ \(\frac{CF}{CD} = \frac{CY}{CX}\) ⇒ \(\frac{CY}{CX} = \frac1n\) Similarly in \(\triangle ABY\), \(\frac{AX}{AY} = \frac{AE}{AB} = \frac1n\) \(\therefore \frac{CY}{CX} = \frac{AX}{AY}\) Also, \(\triangle CFY \cong \triangle AXE\) \(\therefore AX = CY\) \(\frac{AY}{AX} = n\) ⇒ \(\frac{AY + AX}{AX} = n + 1\) ⇒ \(\frac{AY + CY}{AX} = n + 1\) ⇒ \(AX = \frac{AC}{n + 1}\) Now, \(XY = AC - AX - CY\) \(= AC - 2AX\) \(= AC - \frac{AC}{n + 1}\) \(\left(\because AX = \frac{AC}{n + 1}\right)\) \(= \frac{(n + 1) - 1}{n + 1}AC\) \(= \frac{n}{n + 1}a\)

Consider the parallelogram \( A B C D \) as shown in the figure, where \( \frac{A E}{A B}=\frac{C F}{C D}=\frac{1}{n} \), for some positive integer \( n \). Suppose the length of \( A C \) is \( a \), then the length of \( X Y \) is A. \( \frac{a}{n} \). B. \( \frac{n a}{n+1} \). C. \( \frac{(n-1) a}{n+1} \). D. \( \frac{(n-1) a}{n} \).

Answer»

Correct option is (B) \( \frac{na}{n + 1}\)

∵ ABCD is a parallelogram

∴ AB = CD

∴ \(AE = \frac{AB}n = \frac{CD}n = CF\) \(\left(\because \frac{AE}{AB} = \frac{CF}{CD} = \frac 1n (given)\right)\)

In \(\triangle CDX\),

\(\triangle CFY \sim\triangle CDX\)

\(\therefore FY \parallel DX\)

⇒ \(\frac{CF}{CD} = \frac{CY}{CX}\)

⇒ \(\frac{CY}{CX} = \frac1n\)

Similarly in \(\triangle ABY\),

\(\frac{AX}{AY} = \frac{AE}{AB} = \frac1n\)

\(\therefore \frac{CY}{CX} = \frac{AX}{AY}\)

Also,

\(\triangle CFY \cong \triangle AXE\)

\(\therefore AX = CY\)

\(\frac{AY}{AX} = n\)

⇒ \(\frac{AY + AX}{AX} = n + 1\)

⇒ \(\frac{AY + CY}{AX} = n + 1\)

⇒ \(AX = \frac{AC}{n + 1}\)

Now,

\(XY = AC - AX - CY\)

\(= AC - 2AX\)

\(= AC - \frac{AC}{n + 1}\) \(\left(\because AX = \frac{AC}{n + 1}\right)\)

\(= \frac{(n + 1) - 1}{n + 1}AC\)

\(= \frac{n}{n + 1}a\)