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: Strategy to Prove EF ∥ BC:
To prove EF ∥ BC in ΔABC, we use the converse of the Basic Proportionality Theorem (BPT).
For this, we must show:
AE / EB = AF / FC

Step 2: Using Properties of Parallelogram OBGC:
Since OBGC is a parallelogram:
BG ∥ OC
GC ∥ OB

Points O, E, C are collinear.
So BG ∥ OE.

Step 3: Applying BPT in ΔABG:
In triangle ABG,
OE ∥ BG.

By Basic Proportionality Theorem:
AE / EB = AO / OG …(1)

Step 4: Applying BPT in ΔACG:
Since GC ∥ OB and O, F, B are collinear,
GC ∥ OF.

In triangle ACG,
OF ∥ GC.

By Basic Proportionality Theorem:
AF / FC = AO / OG …(2)

Step 5: Comparing (1) and (2):
From (1) and (2):
AE / EB = AO / OG
AF / FC = AO / OG

Therefore,
AE / EB = AF / FC

Step 6: Applying Converse of BPT:
Since the line segment EF divides sides AB and AC in the same ratio,
By converse of Basic Proportionality Theorem:
EF ∥ BC

Final Answer:
EF ∥ BC is proved using BPT in ΔABG and ΔACG and applying the converse of BPT.