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let P be any point on the circumcircle of triangle ABC with radius a then PA2 + PB2+ PC2 =

SUNIL , 9 Years ago
Grade 12
anser 1 Answers
Saurabh Koranglekar

Last Activity: 4 Years ago

Dear student

Let the triangle be right-angled at C and

let the point P coincide at P

PA2+ PB2+ PC2 is (diameter)^2
= 4a^2

Regards

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