Quantitative Aptitude Ques 1118

Question: Let C be a point on a straight line AB. Circles are drawn with diameters AC and AB. Let I be any point on the circumference of the circle with diameter AB. If AP meets the other circle at Q, then

Options:

A) $ QC||PB $

B) $ QC $ is never parallel of $ PB $

C) $ QC=\frac{1}{2}PB $

D) $ QC||PB $ and $ QC=\frac{1}{2}PB $

Show Answer

Answer:

Correct Answer: A

Solution:

  • In $ \Delta AQC $ and $ \Delta APB, $ $ \angle AQC=\angle APB $ [angles made in semi-circle] $ \angle QAC=\angle PAB $ [common]

$ \therefore $ $ \angle ACQ=\angle ABP $

$ \Rightarrow $ $ \Delta AQC\sim \Delta APB $

$ \therefore $ $ \frac{AQ}{AP}=\frac{AC}{AB} $

$ \Rightarrow $ $ QC||PB $