To construct triangle ABC if are given A, h_b + a, h_c + a, where h_b,h_c are the altitudes from B,C, resp.
2Rh_b = ac
2Rh_c = ab
h_b + a = (ac/2R) + a = (a/2R).(2R + c)
h_c + a = (ab/2R) + a = (a/2R).(2R + b)
We have that angle A is known ==> a/2R is known
Let O be the circumcenter and M the midpoint of BC.
The triangle BOC has known angles since A is known ==>
MC / OC is known ==> (a/2)/R = a/2R is known].
So the problem is equivalent to construct triangle if are given A, 2R + b, 2R + c.
Let ABC be the triangle in question, AOD diameter of the circumcircle and B',C' points on the extensions of AB,AC such that BB' = CC' = AD [= 2R].
A is known
AB' = AB + BB' = c + 2R, known
AC' = AC + CC' = b + 2R, known
==> The triangle AB'C' can be constructed.
DB is perpendicular to AB' (since AD is diameter) and BB' = AD [=2R]
==> the locus of D is the parabola with focus A and directrix the perpendicular to AB' at B' (see LEMMA).
DC is perpendicular to AC' and CC' = AD ==> the locus of D is the parabola with focus A and directric the perpendicular to AC' at C'.
Therefore D is intersection point of the two loci. B,C are the (other than A) intersections of the circle of diameter AD with the lines AB',AC' resp.
In triangle ABC, let BC be fixed and D the orthogonal projection of A on BC. If BD = AC then the locus of A is a parabola.
Let A' be the orthogonal projection of A on the perpendicular to BC at B. We have AC = BD and BD = AA' ==> AC = AA' ==> the locus of A is the parabola with focus C and directrix the perpendicular to BC at B.
Let BB' = h_b, CC' = h_c be the altitudes from B,C, resp.
The right triangles C'AC,B'AB are similar and have known angles (since A is known)
CC' / CA = BB' / BA is known ==>
h_c / b = h_b / c = (h_c - h_b) / (b - c) = [(h_c + a) - (h_b + a)] / (b - c)
==> b - c is known.
So the problem is equivalent to construct triangle if are given A, b - c, h_c + a.
Let ABC be the triangle in question with AC > AB.
Let D be the point on AC between A and C such that AD = AB, CE the altitude from C, and Z the intersection of the lines BD and CE.
In the triangle CDZ we have:
CD = AC - AD = b - c, known.
Angles (BDC) = (DAB) + (DBA) = A + (90 - (A/2)) = 90 + (A/2), known
(DCZ) = 90 - (CAE) = 90 - A, known.
Therefore the triangle CDZ can be constructed.
Let H be the point on the extension of CE such that EH = BC.
We have CH = CE + EH = h_c + a, known.
BE is perpendicular to CH and BC = EH ==> (according to LEMMA) the locus of B is the parabola with focus C and directrix the perpendicular to CH at H.
So B is the intersection of the line DZ and the parabola.
Let ABC be the triangle in question and O its circumcenter.
The perpendicular bisector of BC intersects the circumcircle at D,E (as in the figure).
CD = DB =: m
EC = EB =: n
AD := d
EA := e
b + 2R := k1, known (1)
c + 2R := k2, known (2)
m(b + c)= ad (3)(by Ptolemy Theorem in the cyclic quadrilateral ABDC)
The triangle BCD has known angles (DCB = DBC = A/2, CDB = 180 - A)
==> a/m := t is known.
==> b + c = (a/m)d = td
ae + cn = bn (4)(by Ptolemy Theorem in the cyclic quadrilateral ABCE) ==>
e = (b - c)n/a
b - c = k1 - k2, known
n/a is known since the triangle CEB has known angles (CEB = A, ECB = EBC = 90 - (A/2))
Therefore e = (b - c)n/a is known.
EA^2 = AD^2 - AD^2 (5)(by Pythagorean Theorem in the right triangle ADE)or e^2 = 4R^2 - d^2, known.
(1) and (2) ==> b + c = k1 + k2 - 4R (6)
(6) and (3) ==> d = (k1 + k2 - 4R) / t (7)
(7) and (5) ==> 4R^2 - ((k1 + k2 - 4R)/t)^2 = e^2
== > R is known.
So the problem is equivalent to construct triangle if are given A, b - c, R or A, b - c, a (the solution is left to the reader).
To construct triangle ABC if are given:
1. A, h_b - a, h_c + a
2. A, h_b - a, h_c - a