Question

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

• A. Proof Completed
• B. Statement is False
• C. Proof is Incomplete
• D. Cannot be Determined

Hint:

Be clear in your construction description. Draw the perpendicular heights \(DM\) and \(EN\) carefully on your diagram.
This theorem is highly fundamental and serves as the proof foundation for many other properties of triangles in geometry.

Answer 1

The Correct Option is A

Step 1: Understand what is being asked.
We must prove that a line parallel to one side of a triangle, cutting the other two sides, divides them in the same ratio (the Basic Proportionality Theorem). Instead of comparing areas of triangles on the same base, let us prove it using angle-based similarity and a ratio technique called componendo-dividendo.

Step 2: Set up the triangle and given information.
In $\Delta ABC$, let $D$ lie on $AB$ and $E$ lie on $AC$ such that $DE\parallel BC$. We must prove:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
Step 3: Prove $\Delta ADE\sim\Delta ABC$ using AA similarity.
Since $DE\parallel BC$:
$\angle ADE=\angle ABC$ (corresponding angles, as $DE\parallel BC$ with transversal $AB$).
$\angle AED=\angle ACB$ (corresponding angles, as $DE\parallel BC$ with transversal $AC$).
Also, $\angle A$ is common to both triangles.
By AA similarity:
\[ \Delta ADE \sim \Delta ABC \]
Step 4: Write the ratio of corresponding sides.
Since the triangles are similar with $A\to A$, $D\to B$, $E\to C$:
\[ \frac{AD}{AB} = \frac{AE}{AC} \]
Step 5: Apply componendo-dividendo to reach the required ratio.
Take the reciprocal of both sides:
\[ \frac{AB}{AD} = \frac{AC}{AE} \]
Subtract 1 from both sides:
\[ \frac{AB}{AD}-1 = \frac{AC}{AE}-1 \]
\[ \frac{AB-AD}{AD} = \frac{AC-AE}{AE} \]
Since $AB-AD=DB$ and $AC-AE=EC$ (as $D$ lies on $AB$ and $E$ lies on $AC$):
\[ \frac{DB}{AD} = \frac{EC}{AE} \]
Taking the reciprocal once more:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
Final Answer:
The theorem is proved using AA similarity and ratio manipulation, giving $\frac{AD}{DB}=\frac{AE}{EC}$.
\[ \boxed{\frac{AD}{DB}=\frac{AE}{EC}} \]