$\frac{2r(sinA+sinB+sinC)}{4R} = 1-(sin^{2}\frac{B}{2}+sin^{2}\frac{C}{2}) + sin^{2}\frac{A}{2}$$\Leftrightarrow SINa +SINb + SINc = COSc + 2SIN^{2}\frac{a}{2}$
$\Leftrightarrow $ 2R(sinA + sinB + sinC)/4R = 1- [sin²(B/2) + sin²(C/2)] + sin²(A/2)$\Leftrightarrow $ sinA + sinB + sinC = 2sin²(A/2) + cosB + cosC$\Leftrightarrow $ 2sinA/2.cosA/2 + 2sin(B/2+C/2).cos(B/2-C/2) = 2sin²(A/2) + 2cos(B/2+C/2).cos(B/2-C/2) $\Leftrightarrow $ cos(A/2).[cos(B/2+C/2) + cos(B/2-C/2)] = sin(A/2).[cos(B/2+C/2) + cos(B/2-C/2)] $\Leftrightarrow $ 2.cos(A/2).cos(B/2).cos(C/2) = 2sin(A/2).cos(B/2).cos(C/2) (*) B, C là góc tgiác nên 0 < B/2, C/2 < pi/2 => cos(B/2), cos(C/2) > 0 (*) $\Leftrightarrow $ cos(A/2) = sin(A/2) => tan(A/2) = 1 => A/2 = $45^{0}$ => A = $90^{0}$ => $\Delta $ABC vuông tại A