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.
Answer:
- 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. ∴ - 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.