State and prove Basic Proportionality Theorem (Thales Theorem) statement setup:
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.
Step-by-Step Solution
Key Concept: Draw $DE \parallel BC$ in $\Delta ABC$. Express area ratios $\dfrac{\text{ar}(\Delta ADE)}{\text{ar}(\Delta BDE)} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(\Delta ADE)}{\text{ar}(\Delta DEC)} = \dfrac{AE}{EC}$.
Given: $\Delta ABC$ in which line $DE \parallel BC$ intersects $AB$ at $D$ and $AC$ at $E$. To prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. [0.5 Mark]
Construction: Join $BE, CD$. Draw $EN \perp AB$ and $DM \perp AC$. [0.5 Mark]
Proof: $\text{ar}(\Delta ADE) = \dfrac{1}{2} \times AD \times EN$ and $\text{ar}(\Delta BDE) = \dfrac{1}{2} \times DB \times EN \Rightarrow \dfrac{\text{ar}(\Delta ADE)}{\text{ar}(\Delta BDE)} = \dfrac{AD}{DB}$ -- (1).
Similarly, $\dfrac{\text{ar}(\Delta ADE)}{\text{ar}(\Delta DEC)} = \dfrac{AE}{EC}$ -- (2). [0.5 Mark]
Since $\Delta BDE$ and $\Delta DEC$ lie on the same base $DE$ and between same parallels $DE \parallel BC$, $\text{ar}(\Delta BDE) = \text{ar}(\Delta DEC)$.
From (1) and (2), $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Given, To Prove, and Construction: 0.5 Mark
Area ratio equations (1) and (2): 0.5 Mark
Equal areas of triangles on same base and parallel lines: 0.5 Mark
Final conclusion $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: 0.5 Mark
Correct Answer: