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.   AP=PQ(i)[ By intercept theorem ]
  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.   PQ=QC(ii)[ By intercept theorem ] Thus, from eq (i) and eq (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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