Question

State and prove Basic Proportionality Theorem.

Hint:

If two triangles are similar, the ratio of any two corresponding segments (medians, altitudes, bisectors) is equal to the ratio of their corresponding sides.

Answer 1

Step 1: Understanding the Concept:
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.
Step 2: Key Formula or Approach:
Ratio of areas of triangles with the same height is the ratio of their bases.
Step 3: Detailed Explanation:
1. Given: \(\triangle ABC\), \(DE \parallel BC\). 2. To prove: \(AD/DB = AE/EC\). 3. Construction: Join \(BE\) and \(CD\). Draw \(DM \perp AC\) and \(EN \perp AB\). 4. Proof: Area(\(\triangle ADE\)) / Area(\(\triangle BDE\)) = \(\frac{1}{2} \cdot AD \cdot EN / \frac{1}{2} \cdot DB \cdot EN = AD/DB\). Area(\(\triangle ADE\)) / Area(\(\triangle CED\)) = \(\frac{1}{2} \cdot AE \cdot DM / \frac{1}{2} \cdot EC \cdot DM = AE/EC\). 5. Since \(\triangle BDE\) and \(\triangle CED\) are on the same base \(DE\) and between the same parallels \(DE\) and \(BC\), their areas are equal. 6. Therefore, \(AD/DB = AE/EC\).
Step 4: Final Answer:
The theorem is proved using the area-ratio property.