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 Statement (Basic Proportionality Theorem):
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the two sides are divided in the same ratio.

Given: In ΔABC, DE ∥ BC where D lies on AB and E lies on AC.
To Prove: AD / DB = AE / EC

Step 2: Construction:
Join BE and CD.
Draw DM ⟂ AC and EN ⟂ AB.

Step 3: Using Area Concept:
Consider triangles ADE and BDE.

Area(ΔADE) = (1/2) × AD × EN
Area(ΔBDE) = (1/2) × DB × EN

Therefore,
Area(ΔADE) / Area(ΔBDE) = AD / DB → (1)

Now consider triangles ADE and CED.

Area(ΔADE) = (1/2) × AE × DM
Area(ΔCED) = (1/2) × EC × DM

Therefore,
Area(ΔADE) / Area(ΔCED) = AE / EC → (2)

Step 4: Key Observation:
Triangles BDE and CED are on the same base DE and between the same parallels DE and BC.
Hence,
Area(ΔBDE) = Area(ΔCED)

Step 5: Final Deduction:
Since Area(ΔBDE) = Area(ΔCED),
From (1) and (2),
AD / DB = AE / EC

Final Conclusion:
Thus, when a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
Hence, the theorem is proved using the area-ratio property.