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Triangles
CBSE 2026 Board Exam Set 3 (Code 30/1/3)
CBSE_BOARD_PYQ_2026_30_1_3
Grade 10
Question:
[Section D]
State and prove Basic Proportionality Theorem (Thales Theorem).
State and prove Basic Proportionality Theorem (Thales Theorem).
Step-by-Step Solution
Key Concept: Statement + Construction + Proof using ratios of areas of triangles.
Statement: 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. [1.0 Mark]
Proof: Draw $EN \perp AB$ and $DM \perp AC$. Join $BE$ and $CD$. $\text{Area}(\Delta ADE) = \frac{1}{2} AD \cdot EN$, $\text{Area}(\Delta BDE) = \frac{1}{2} DB \cdot EN \Rightarrow \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{AD}{DB}$. Similarly, $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta DEC)} = \frac{AE}{EC}$. Since $\Delta BDE$ and $\Delta DEC$ are on same base $DE$ and between same parallels $DE \parallel BC$, $\text{Area}(\Delta BDE) = \text{Area}(\Delta DEC) \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$. [4.0 Marks]
Proof: Draw $EN \perp AB$ and $DM \perp AC$. Join $BE$ and $CD$. $\text{Area}(\Delta ADE) = \frac{1}{2} AD \cdot EN$, $\text{Area}(\Delta BDE) = \frac{1}{2} DB \cdot EN \Rightarrow \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{AD}{DB}$. Similarly, $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta DEC)} = \frac{AE}{EC}$. Since $\Delta BDE$ and $\Delta DEC$ are on same base $DE$ and between same parallels $DE \parallel BC$, $\text{Area}(\Delta BDE) = \text{Area}(\Delta DEC) \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$. [4.0 Marks]
Correct Answer: Proof completed.
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