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Question
P and Q are points on the sides AB and AC respectively of a ΔABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm, and QC = 6 cm, show that BC = 3PQ.
P and Q are points on the sides AB and AC respectively of a ΔABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm, and QC = 6 cm, show that BC = 3PQ.
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Solution
AP/PB = AQ/QC
2/4 =3/6
1/2 = 1/2
Therefore, PQ || BC [Converse of basic proportionality theorem.]
ΔAPQ ~ ΔABC
AP/AB = AQ/AC = PQ/BC
2/6 = 3/9 = PQ/BC
PQ/BC = 1/3
BC = 3 PQ.
Hence proved
AP/PB = AQ/QC
2/4 =3/6
1/2 = 1/2
Therefore, PQ || BC [Converse of basic proportionality theorem.]
ΔAPQ ~ ΔABC
AP/AB = AQ/AC = PQ/BC
2/6 = 3/9 = PQ/BC
PQ/BC = 1/3
BC = 3 PQ.
Hence proved
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