AP, BQ, CR r the diameters of the circumcircle of triangle ABC. Let a1, a2, a3, a be areas of triangles BPC, CQA, ARB and ABC respectively. Then i) a1=ii) a1+ a2+a3,=iii) BP2 + CQ2+ AR2 =

Dear

BP² =AP² -AB²

     =4R² -c²

similerly

CP² =AP² -AC²

     =4R² -b²

We know  Δ =abc/4R

trianble BPC also have same circumradius

so a1=BC*CP*PB/4R

        =a V(4R² -b²) V(4R² -c²) /4R

similerly you can find a2 and a2

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