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Question
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 in P and Q respectively. Prove that P and Q divide AC into three equal parts.
Solution
Draw DE || BC through A
AM = MN = NB ...(given)
MP || BC ; NQ || BC ...(given)
DE || BC
i.e. AM, MN and NB are equal intercepts made on transversal AB.
AC is also a transversal; intercepts made on AC are AP, PQ and QC.
Hence, AP = PQ = QC
Therefore, P and Q divide AC in three equal parts.
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