Dear Luis,

[LL]: Both these problems have easy algebraic solutions/constructions.

One can easily find b and c from b+c=s ; bc=u^2 and b-c=d ; bc=v^2.

But... I am looking for an elegant "pure" geometrical approach. Is

there one?

*** Here is one. The relation (b+c)^2 + (b-c)^2 = 4m_a^2 + a^2

shows that the two problems can be solved simultaneously.

Given a segment $BC$ of length a, midpoint D, and a perpendicular

segment A_0D = m_a, construct

(i) a semicircle center D through A_0,

(ii) the circle with diameter A_0B (with perpendicular diameter MN, M

and C on opposite sides of A_0B),

(iii) the circle, center M, through A_0 and B.

Let P be a point on (ii) with BP = b + c for P on the semicircle A_0MB

and BP=b-c for P on A_0NB.

Extend BP to intersect the circle (iii) at Q.

Construct the circle through Q with center B to intersect the

semicircle (i) at A.

Then ABC is the required triangle.

Best regards

Sincerely

Paul