Question: Let C be a point on a straight line AB. Circles are drawn with diameters AC and AB. Let P 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 $