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Question

Let Δ ABC be an equilateral triangle. Let P be a point on minor arc AB of its circumcircle . Which of the following holds :


A
PC=2PB
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B
PC2=PA2+PB2
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C
PA=PB+PC
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D
PC=PA+PB
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Solution

The correct option is B PC=PA+PB

Draw PA,PB,PC. By Ptolemy's Theorem applied to quadrilateral APBC,

we know that PC×AB=PA×PB+PB×AC.

Since,AB=BC=CA=s, we divide both sides of the last equation by to get the result:PC=PA+PB.


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