Question

In the given \(∆ABC\), if \(DE || BC, AE = a\) units, \(EC = b\) units, \(DE = x \) units and \(BC = y\) units, then which of the following is true?

• A. \(x=\frac{ay}{a+b}\)
• B. \(y=\frac{ax}{a+b}\)
• C. \(x=\frac{a+b}{ay}\)
• D. \(\frac{x}{y}=\frac{a}{b}\)

Answer 1

The Correct Option is A

Given: In \( \triangle ABC \), \( DE \parallel BC \), and the given segment lengths: 

• \( AE = a \)
• \( EC = b \)
• \( DE = x \)
• \( BC = y \)

Step 1: Apply Basic Proportionality Theorem (Thales' Theorem)

Since \( DE \parallel BC \), by the theorem:

\[ \frac{DE}{BC} = \frac{AE}{AC} \] \[ \frac{x}{y} = \frac{a}{a+b} \]

Step 2: Solve for \( x \)

\[ x = \frac{ay}{a+b} \]

Final Answer: \( x = \frac{ay}{a+b} \)

Answer 2

In \(∆ABC\), given that \(DE || BC\), we can apply the Basic Proportionality Theorem (also known as Thales' theorem). This theorem states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those two sides proportionally.

So, given \(DE || BC\), we know that:

\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Given that \(AE = a\) units and \(EC = b\) units, we can substitute these in:

\[ \frac{AD}{DB} = \frac{a}{b} \]

Furthermore, since \(DE\) is parallel to \(BC\), the segments \((AD + DB)\) and \((AE + EC)\) are proportional to the segments \(DE\) and \(BC\), respectively. Thus:

\[ \frac{DE}{BC} = \frac{AE}{AC} \]

Since \(AC = AE + EC = a + b\), we get:

\[ \frac{x}{y} = \frac{a}{a+b} \]

Therefore, solving for \(x\), we have:

\[ x = \frac{a \cdot y}{a + b} \]

This solution aligns with the first option: \(x = \frac{ay}{a+b}\).