Question

In the same figure as 32(a), prove that \(EF \parallel BC\).

Hint:

Whenever you need to prove lines are parallel in a triangle, look for a common ratio using BPT. Here, \(AO/OG\) acts as the bridge between the two sides.

Answer 1

Step 1: Understanding the Concept:
To prove \(EF \parallel BC\), we can use the converse of Basic Proportionality Theorem (BPT) in \(\triangle ABC\).
We need to show that \(\frac{AE}{EB} = \frac{AF}{FC}\).
Step 2: Detailed Explanation:
1. From part (i), we know \(OBGC\) is a parallelogram.
Therefore, \(BG \parallel OC\) and \(GC \parallel OB\).
2. Since \(BG \parallel OC\) and \(O, E, C\) are collinear, we have \(BG \parallel OE\).
In \(\triangle ABG\), \(OE\) is a line segment parallel to the base \(BG\).
By Basic Proportionality Theorem (BPT):
\[ \frac{AE}{EB} = \frac{AO}{OG} \quad \dots(1) \]
3. Similarly, since \(GC \parallel OB\) and \(O, F, B\) are collinear, we have \(GC \parallel OF\).
In \(\triangle ACG\), \(OF\) is parallel to the base \(GC\).
By BPT:
\[ \frac{AF}{FC} = \frac{AO}{OG} \quad \dots(2) \]
4. From equations (1) and (2), we equate the ratios:
\[ \frac{AE}{EB} = \frac{AF}{FC} \]
5. In \(\triangle ABC\), since the line segment \(EF\) divides sides \(AB\) and \(AC\) in the same ratio, by the converse of BPT:
\[ EF \parallel BC \]
Step 3: Final Answer:
Using BPT in \(\triangle ABG\) and \(\triangle ACG\), we established \(\frac{AE}{EB} = \frac{AF}{FC}\), which proves \(EF \parallel BC\) by the converse of BPT.