In the given figure, points M and N divide the side AB of ΔABC into three equal parts. Line segments MP and NQ are both parallel to BC and meet AC at P and Q respectively. Prove that P and Q divide AC into three equal parts i.e AC=3AP=3PQ=3QC.

A B C X Y M N P Q


Answer:


Step by Step Explanation:
  1. Through A, let us draw XAY||BC.

    Now, XY||MP||NQ are cut by the transversal AB at A,M, and N respectively such that AM=MN.

    Also, XY||MP||NQ are cut by the transversal AC at A,P, and Q respectively.
  2. Again, MP || NQ || BC are cut by the transversal AB at M, N, and B respectively such that MN = NB.

    Also, MP || NQ || BC are cut by the transversal AC at P, Q, and C respectively. \begin{aligned} \therefore \space\space PQ = QC &&\ldots\text{(ii)} &&[\text{ By intercept theorem }] \end{aligned} Thus, from eq \space (\text{i}) and eq \space (\text{ii}), we get AP = PQ = QC.

    Therefore, P and Q divide AC into three equal parts.

    Hence, AC = 3AP = 3PQ = 3QC.

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