Question

State and prove Basic Proportionality Theorem.

Hint:

While proving BPT, remember that the altitude for an obtuse-angled triangle (like $\Delta BDE$) lies outside the triangle. This is why $EN$ is the height for both $\Delta ADE$ and $\Delta BDE$.

Answer 1

Step 1: Understanding the Concept:
The Basic Proportionality Theorem (BPT), also known as Thales Theorem, states 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.
Step 2: Key Formula or Approach:
We use the concept of the area of triangles.
Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$.
If two triangles have the same base and lie between the same parallel lines, their areas are equal.
Step 3: Detailed Explanation:
Given: In $\Delta ABC$, a line $DE$ is parallel to $BC$ ($DE \parallel BC$), intersecting $AB$ at $D$ and $AC$ at $E$.
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$
Construction: Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.
Proof:
Calculate Area($\Delta ADE$) taking $AD$ as base:
\[ \text{Area}(\Delta ADE) = \frac{1}{2} \times AD \times EN \]
Calculate Area($\Delta BDE$) taking $DB$ as base:
\[ \text{Area}(\Delta BDE) = \frac{1}{2} \times DB \times EN \]
Divide these two areas:
\[ \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \dots(1) \]
Similarly, calculate Area($\Delta ADE$) taking $AE$ as base and Area($\Delta CDE$) taking $EC$ as base:
\[ \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \dots(2) \]
Notice that $\Delta BDE$ and $\Delta CDE$ are on the same base $DE$ and between the same parallels $DE$ and $BC$.
Therefore, Area($\Delta BDE$) = Area($\Delta CDE$) $\dots(3)$.
From (1), (2), and (3), we can conclude that the ratios of the areas are equal.
Step 4: Final Answer:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
Hence proved.