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Question

Square ABCD is inscribed in a circle, and P is a point on arc BC of the circle.


A
PA+PCPB+PD = PDPA
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B
PD+PCPB+PA = PDPA
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C
PA+PCPB+PD = PAPD
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D
None of the Above.
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Solution

The correct option is B PA+PCPB+PD = PDPA

By applying Ptolemy's Theorem to cyclic quadrilaterals PBAD and PADC respectively we have:
PA×BD=PB×AD+PD×AB=AB(PB+PD)
PD×AC=PC×AD+PA×CD=AB(PA+PC)
Now by dividing these two we get,

PA+PCPB+PD = PDPA


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