Question

In ΔABC, PQ || BC. Prove that

1. Area (ΔPBC) = Area (ΔQBC)
2. Area (ΔPOB) = Area (ΔQOC)

Answer 1

Given: PQ || BC

Step 1: Use the property of parallel lines in triangles

If a line is drawn parallel to the base of a triangle, then it divides the other two sides proportionally:

`(AP)/(AB) = (AQ)/(AC)`

i. Prove that Area(ΔPBC) = Area(ΔQBC)

• Let h be the perpendicular distance from B to PQ and h’ be the perpendicular distance from C to PQ
• Triangles ΔPBC and ΔQBC share the same base BC.

The area of a triangle:

Area = `1/2` × base × height

• For ΔPBC, height = perpendicular from P to BC (or along the line joining P parallel to BC) and similarly for ΔQBC.
• Since PQ || BC, the perpendicular distances from P and Q to BC are the same.
  Area(ΔPBC) = `1/2` × BC × height = Area(ΔQBC)

Hence, Area(ΔPBC) = Area(ΔQBC).

ii. Prove that Area(ΔPOB) = Area(ΔQOC)

Assume O is the point of intersection of the medians or some reference point inside the triangle (typically the centroid or origin).

• Triangles ΔPOB and ΔQOC share the same height from line PQ to O because PQ || BC.
• Also, the bases PB and QC are equal due to the proportional division by the parallel line:
  `(AP)/(AB) = (AQ)/(AC)`
  ⇒ PB = QC
• Using the area formula for a triangle:
  Area = `1/2` × base × height
  Area(ΔPOB) = Area(ΔQOC)

Hence, Area(ΔPOB) = Area(ΔQOC).