The many ways to construct a triangle II

h_c, l_c, m_c

a, b, A

This is a well known construction and, in principle, the three elements a, b, and A may not define a triangle or may not define it uniquely. When the angle is right the triangle is defined in a unique way.

Draw AC of the given length b and the angle A. The circle of radius a and centered at C may intersect the other leg of A either at two points B', B", at one point B (it will actually touch AB) or not at all. When the point of intersection is unique the angle B is right. If this is known in advance, we may use a different approach.

h_a, b, c

Draw two parallel lines h_a (units) apart. Pick a point A on one of them and from this point and with the center at A draw arcs of radii a and b till they intersect the other line. The problem may have no solution (e.g., b<h_a or b<h_a). Otherwise, it may have one solution (b=h_a or c=h_a). And lastly, it may have two solutions ABC' and ABC" as indicated in the diagram.

aa, H_b, H_c

Triangles BH_bC and BH_cC are both right. Therefore, the two given points H_b and H_c lie on the circle with diameter BC. The perpendicular bisector of the segment H_bH_c passes through the center M of this circle. This suggests the following construction:

Find the point M at the intersection of aa and the perpendicular bisector of H_bH_c. Draw a circle centered at M through either H_b or H_c. Extend BH_c and CH_b until they intersect at A.

Second Construction (Barukh Ziv)

a, h_b, l_c

Draw an auxiliary triangle BCH_b and the bisector of the angle C. On this bisector, select L_c so that CL_c = l_c. Extend BL_c and CH_b until they meet at A.

h_a, h_b, c

Draw an auxiliary triangle ABH_b and the circle with radius h_a centered at A. Draw a tangent from B to the circle. Extend the tangent and AH_b until they meet at C.

h_a, h_b, m_c

Assume the triangle ABC has been constructed. From M_c drop a line M_cH' perpendicular to CB. Then M_cH' is half h_a. This suggests the following construction:

First construct the right triangle CM_cH' with M_cH' = h_a/2 and hypotenuse CM_c = m_c. CH' defines the line aa. Similarly, on the other side of CM_c find the line bb. Extend CM_c to twice its length to get the point D from which draw lines parallel to aa and bb to obtain A and B, respectively.