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Bhavya

· started a discussion

· 1 Months ago

Wrong solution (Question: 108529.0 )

Option D will be correct. Using the property of similar triangle QC will be half of PB as AC(radius of bigger circle) is half of AB(diameter of bigger circle)

Question:
Assume C is a point on a straight line AB and two circles of diameter AC and AB are drawn on this line. P is a point on circumference of the circle of diameter AB. If AP meets second circle at Q, then which of the following is correct?
Options:
A) QC || PB
B) QC will never be parallel to PB
C) QC = \(\cfrac{1}{2}\)PB
D) QC || PB and QC = \(\cfrac{1}{2}\)PB
Solution:
Ans: (a)

From figure,


\(\angle \)APB = \(\angle \)AQC and \(\angle \)PAB = \(\angle \)QAC

DPAB ~ DQAC

\(\therefore\) \(\angle \)PBA = \(\angle \)QCA

Hence, QC | | PB.

shailesh raj

· commented

· 1 Months ago

here midpoint theorem cannot be used... because here any midpoint or half is not given and you cannot assume it by yourself.
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