To construct triangle ABC if are given A, a, h_a + h_b + h_c (sum of altitudes)
A and a known ==> R is known.
h_a + h_b + h_c = 2R(bc + ca + ab)
==> bc + ca + ab is known.
bc + ca + ab = a(b+c) + bc := k^2 (known)
a^2 = b^2 + c^2 - 2bc.cosA = (b+c)^2 - 2bc(1 + cosA) = (b+c)^2 - 4bc(cos(A/2))^2
Denote b+c := X, bc := Y^2
aX + Y^2 = k^2
a^2 = X^2 - 4Y^2(cos(A/2)^2
==> X^2 + 4a(cos(A/2)^2.X - 4k^2(cos(A/2)^2 - a^2 = 0
==> b+c is known and also bc is known.
==> b,c are known.
2(bc + ca + ab) = (b + c)^2 - (b^2 + c^2) + 2a(b + c) (1)
Let ABC be the triangle in question. The bisector AD of A intersects the circumcircle at E. Let EF be the diameter perpendicular to BC at its midpoint M (see figure).
AE = d, AM = m_a,
EB = EC = m, known
AF = y
EM = x, MF = 2R - x = z, known
(since the isosceles triangles EBC, FBC are known: BC = a and have known angles.)
y^2 = (2R)^2 - d^2 (from the right triangle AEF)) (2)
m(b + c) = ad (by Ptolemy Theorem in the cyclic quadril. ABEC)
==> b + c = am / d (3)
b^2 + c^2 = 2(m_a)^2 + (a^2 / 2) (Theorem of median in ABC) (4)
zd^2 + xy^2 = 2R(m_a)^2 + 2Rxz (by Stewart Theorem in AFE) (5)
(2) and (5) ==> zd^2 + x((2R)^2 - d^2) = 2R(m_a)^2 + 2Rxz (6)
(1) and (3), (4), (6) ==>
2(bc + ca + ab) = (a/m)^2d^2 - ((z-x)/R)d^2 - 4Rx + 2xz - (a^2)/2 + (2a^2/m)d
[(z-x)/R - (a/m)^2]d^2 - (2a^2/m)d - 2xz + 4Rx + (a^2/2) + 2(bc + ca + ab) = 0
==> AE = d is known.
We construct the isosceles triangle EBC with BC = a, BEC = 180 - A, BE = CE. The circle (E, d) intersects the circumcircle of EBC at A.