question_answer

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            [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]

\[\therefore \]      \[\angle ACQ=\angle ABP\]

\[\Rightarrow \]   \[\Delta AQC\sim \Delta APB\]

\[\therefore \]      \[\frac{AQ}{AP}=\frac{AC}{AB}\]\[\Rightarrow \]\[QC||PB\]