m_a,m_b,b
For the same reason, this is a SSS problem with side lengths (2m_a/3 , b/2 , m_b/3)

R,h_a,a
Draw an isosceles triangle with side lengths (R,R,a) and vertex O. Cut the circumcircle with a line parallel to a at distance h_a.

R,m_a,a
Draw an isosceles triangle with side lengths (R,R,a) and vertex O. Draw a circle with center in the midpoint of a and radius m_a and intersect it with the circumcircle.

a,h_a,m_a
Draw a line parallel to a at distance h_a, then intersect it with a circle centered in the midpoint of a having radius equal to m_a.

A,a,m_a
Draw the circle that sees a under an angle equal to A. Intersect it with a circle centered in the midpoint of a having radius m_a.

A,B,h_c
Draw the altitude h_c and a line l perpendicular to it. Draw the circles that see h_c under angles equal to B,C and intersect them with l.

A,h_a,l_a
Draw a right triangle having side lengths h_a, l_a, sqrt(l_a^2-h_a^2). Intersect the last side with the lines forming angles equal to +A/2 ,-A/2 with l_a.

A,a,r
The distance between I and a is r, the angle ^BIC is (pi+A)/2, so we can solve the problem by intersecting a circle and a line parallel to a.

A,a,R
Is not a base.

m_a,h_a,h_b
We draw a circle having m_a as diameter, then intersect it with two circles centered in the endpoints of m_a and having radii equal to h_a, h_b/2. In this way we determine the feet H_A of the altitude h_a and the direction of the side b, so we find the vertex C, then the vertex B.

m_a,h_b,h_c
We draw two opposite right triangles sharing a common hypotenuse m_a and having sides opposite to vertex A with lengths h_b/2, h_c/2. In this way we determine the lines containing the sides (b,c), having only to find two points on these lines with the proper distances (h_c,h_b) from (b,c).

m_a,h_a,m_b
We draw a right triangle having side lengths (h_a,m_a,sqrt(m_a^2-h_a^2)). We call l the line containing the last side, and G the point on m_a such that 3 AG = m_a. We intersect a circle centered in G with radius 2m_b/3 with the line l, finding B, then C.

m_a,m_b,h_c
We glue two right triangles such that the triangle GAB satisfies
GA = 2m_a/3
GB = 2m_b/3
d(G,AB) = h_c/3
then call (U,V) the midpoints of (GA,GB). We call M_a the symmetric
of U with respect to G, and M_b the symmetric of V with respect to G.
AM_b and BM_a concur in C.

a,h_b,R
We draw an isosceles triangle with side lengths (R,R,a) and vertex O.
We draw a circle having a as a diameter, and find a point H_B on it'such that BH_B = h_b. The intersection of the line CH_B with the circumcircle is A.

a,h_b,m_a
We build a triangle BCH_B having BH_B perpendicular to CH_B, BH_B=h_b,
BC=a. We draw a circle centered in the midpoint of BC having radius m_a, the intersect it with the line CH_B, finding A.

h_a,l_a,b
We build a triangle ACH_A having AC=b, AH_A=h_a, CH_A perpendicular
to AH_A. On the line CH_A we determine L_A such that AL_A=l_a. We call
D the symmetric of C with respect to the line AL_A. The intersection of AD and CL_A is B.

A,h_a,h_b
We build a right triangle ABH_B having AH_B=h_b, ^BAH_B=A, AH_B
perpendicular to BH_B. We find a point H_A such that BH_A is perpendicular to AH_A and AH_A=h_a by drawing the circle having AB as a diameter. The intersection of BH_A and AH_B is C.

a,m_a,h_a
We draw a line l parallel to a such that d(a,l)=h_a and intersect it with the circle having radius m_a and center in the midpoint of a.

(a,m_a,l_a)(a,h_a,l_a)
Still thinking about.

m_a,h_a,l_a (Own)
We draw M_A,H_A and L_A such that they lie on a line l perpendicular to AH_A and satisfy AH_A = h_a, AM_A=m_a, AL_A=l_a. We intersect the perpendicular to l through M_A (the axis of BC) with the line AL_A, finding E, that is the midpoint of the arc BC in the circumcircle
(little folklore apart: in the Italy IMO team this fact is generally called "Tiozzo's lemma", from the great contestant Giulio Tiozzo).
The intersection of the axis of AE with the line EM_A is O, the circumcenter. So we draw the circumcircle with radius OA and find B and C on l.

We draw AL_A such that AL_A=l_a, then the circle having
l_a as a diameter. On this circle we find a point H_A such
that AH_A=h_a. Now the angle ^H_A A L_A = (B-C)/2, and
we fall into another famous construction:

(a,h_a,B-C)

Here we exploit this fact: if two triangles ABC and DBC are
symmetric wrt to the axis of BC, they share the same circumcircle
and satisfy ^ABD = (B-C).

So we draw BC such that BC=a and find a point F on the axis
of BC such that d(F,a)=h_a. Now we draw a segment UV, parallel
to BC, with midpoint F, then we draw the circle Gamma, locus
of points G for which ^UGV=(B-C).

Now we exploit homotethy: by calling X the intersection of
Gamma and BF, we have that the parallel to UX through B
and the line UV meet on A, solving the problem.

(Thanks to Federico Stra for the brilliant idea.)

Now, back to the original problem:

(a,h_a,l_a)

We draw BC=a with midpoint M, then H such that HM=h_a
lies on the axis of BC. We draw a line r through H, parallel to BC,
then find L on it'such that LM=l_a. We call T the intersection
of BH with the circumcircle of MHL. The line HL and the parallel
to TL through B meet on A.

(A,a,b+c)
Conway solution: draw a triangle DBC with
^BDC=A/2, CD=b+c, BC=a
then intersect the perpendicular bisector of DB
with CD, finding A. DAB is isosceles in A
and ^CAB=A holds.

Jack solution: draw two rays (r,s)
forming an angle equal to A, and pick
K on r such that AK=(a+b+c)/2, J on S
such that AJ=(b+c-a)/2. Call I and I_A
the intersection of the perpendicular
to s through J and the perpendicular to
r through K with the bisector of (r,s).

Intersect the circle having II_A as a diameter
with r and s, finding B and C.
center I and radius IJ.]

(A,b,a+c)
Conway style: we draw CAD such that AC=b, ^CAD=A,
AD=a+c, then intersect the perpendicular bisector
of CD with AD, finding B.

Jack style: we draw CAD such that AC=b, ^CAD=a,
AD=p, then call I_A the intersection of the bisector
of A with the perpendicular to AD through D.
We draw l_C, the perpendicular to CI_A through C,
then reflect AC wrt l_C, finding CB.

(A,a,b-c)
We draw BC=a with J on it'such that 2 BJ = a-(b-c).
On the perpendicular to BC through J we find I
such that 2 ^BIC = pi+A, then reflect the line
BC wrt BI and CI, finding A.

Instructive constructions for
(a,m_a,l_a)
and
(A,a,l_a)
are due to Paul Yiu and appear in:
https://forumgeom.fau.edu/FG2005volume5/FG200512.pdf

The case
(h_a,l_a,m_b)
was solved by Gilles Boutte and can be found here:
https://g.boutte.free.fr/articles/021212.pdf

PS: https://www.cut-the-knot.org/Curriculum/Geometry/Kung.pdf
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