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. ∴ AP=PQ…(i)[ By intercept theorem ] - 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.