| 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 [SSC (CGL) 2014] |
A) \[QC||PB\]
B) \[QC\] is never parallel of \[PB\]
C) \[QC=\frac{1}{2}PB\]
D) \[QC||PB\]and \[QC=\frac{1}{2}PB\]
Correct Answer: A
Solution :
| In \[\Delta AQC\]and \[\Delta APB,\] |
| \[\angle AQC=\angle APB\] |
| [angles made in semi-circle] |
 |
| \[\angle QAC=\angle PAB\][common] |
| \[\therefore \] \[\angle ACQ=\angle ABP\] |
| \[\Rightarrow \] \[\Delta AQC\sim \Delta APB\] |
| \[\therefore \] \[\frac{AQ}{AP}=\frac{AC}{AB}\] |
| \[\Rightarrow \] \[QC||PB\] |
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