A) PQ
B) 2PQ
C) 3PQ
D) 4PQ
Correct Answer: C
Solution :
Since, \[\frac{AP}{PB}=\frac{AQ}{QC}\] \[\therefore \]\[PQ||BC,\] [By converse of Thale's theorem] \[\Rightarrow \]\[\angle APQ=\angle ABC\] and \[\angle AQP=\angle ACB\] (Corresponding angles) \[\therefore \] \[\Delta APQ\tilde{\ }\Delta ABC\] [By AA similarity] \[\Rightarrow \] \[\frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC}\] \[\Rightarrow \] \[\frac{PQ}{BC}=\frac{1}{3}\] \[\Rightarrow \] \[BC=3PQ\]You need to login to perform this action.
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