Question

In a trapezium ABCD, AB || CD. M and N are two points on AD dividing it into three equal parts. Line segments MP and NQ are parallel to AB which meet BC at P and Q, respectively. Prove that BP = PQ = CQ, i.e., P and Q divide BC into three equal parts.

Answer 1

Given: In trapezium ABCD, AB || CD.

Points M and N lie on AD such that AM = MN = ND (AD is divided into three equal parts).

Lines MP and NQ are drawn parallel to AB, meeting BC at P and Q respectively.

To Prove: BP = PQ = QC i.e., P and Q divide BC into three equal parts.

Proof (Step-wise):

1. From the hypothesis MP || AB and NQ || AB and given AB || CD, we have four parallel lines:

AB || MP || NQ || CD

2. Consider the transversal AD.

The four parallel lines meet AD at A, M, N and D respectively.

By construction AM = MN = ND (AD is divided into three equal parts).

3. Consider the other transversal BC.

The same four parallel lines meet BC at B, P, Q and C respectively.

4. By the Intercept Theorem if a set of parallel lines cut one transversal in equal segments, they cut any other transversal in equal corresponding segments, equal intercepts on AD (AM = MN = ND) imply equal intercepts on BC: BP = PQ = QC.

Hence, BP = PQ = QC. 

So, P and Q divide BC into three equal parts.