Geometry: Converse of Basic Proportionality Theorem

 Converse of Basic Proportionality Theorem Theorem:(Converse of Basic Proportionality Theorem) If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
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Example: ABCD is a quadrilateral; P, Q, R and S are the points of trisection of sides AB, BC, CD and DA respectively and are adjacent to A and C; prove that PQRS is a parallelogram.
Solution: A quadrilateral ABCD in which P, Q, R and S are the points of trisection of sides AB, BC, CD and DA respectively and are adjacent to A and C. ...[Given]

You have to prove, PQRS is a parallelogram i.e. PQ SR and QR PS

You have to made a construction by joining AC.

Since P, Q, R and S are the points of trisection of AB, BC, CD and DA respectively.
 BP = 2 PA, BQ = 2 QC, DR = 2 RC and DS = 2 SA. In ADC, you have DS/ SA = 2SA/ SA = 2 and, DR/ RC = 2RC/RC = 2 => DS/ SA = DR/ RC => S and R divide the sides DA and DC respectively in the same ratio. => SR AC ...(i) [By the converse of Thale's Theorem] In ABC, you have BP/PA = 2PA/PA = 2 and BQ/QC = 2QC/QC = 2 => BP/ PA = BQ/QC => P and Q divide the sides BA and BC respectively in the same ratio. => PQ AC ...(ii) [By the converse of Thale's Theorem]

From equations (i) and (ii), you have
SR AC and PQAC
SR PQ

Similarly, by joining BD, you can prove that
QR PS.
Hence, PQRS is a parallelogram.

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