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