Question

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Hint:

While proving theorems, always provide a neat diagram and clearly label "Given", "To Prove", and "Construction".

Answer 1

Theorem:
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Given:
In ΔABC, a line l is drawn parallel to BC, meeting AB at D and AC at E.
So, DE ∥ BC.

To Prove:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Construction:
Join B to E and C to D to form triangles ADE and ABC.

Proof:
Step 1: Since DE ∥ BC, ∠ADE = ∠ABC (corresponding angles).
Also, ∠AED = ∠ACB (corresponding angles).

Step 2: Therefore, ΔADE ∼ ΔABC (by AA similarity criterion).

Step 3: For similar triangles, corresponding sides are proportional:
\[ \frac{AD}{AB} = \frac{AE}{AC} \]
Rewrite AB and AC in terms of segments:
\[ AB = AD + DB,\quad AC = AE + EC \]
So, \[ \frac{AD}{AD + DB} = \frac{AE}{AE + EC} \]

Step 4: Use property of equal fractions:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Final Conclusion:
A line drawn parallel to one side of a triangle divides the other two sides in the same ratio.

Hence proved.