i) of side a=|BC|, ii) of altitude h

The clue is to consider the symmetric B' of B on the parallel to side BC. Triangle B'BC is constructible from the data. The same is true for triangle ADE. Hence angle(DAE) = B-C (see Bisector.html ) is constructible from the data.

Besides, for F on the extension of AC: angle(FAB') = angle(BFA)-angle(FB'A) = (pi/2-C)-(pi/2-B) = B-C.

Thus A is intersection point of the medial line of BB' and the arc on B'C viewing it under the angle w=pi-(B-C).

i) of side a=|BC|, ii) of median m

Use the formulas expressing the lengths as functions of the sides (see Stewart.html ).

To solve the system for b, c, set b+c=x and bc=y

Solution proposed by Fursenko in his remarkable exposition of

i) of angle A, ii) of median m

Solution after G. Velissarios (AMM 1988, p. 458). Assume the triangle constructed and take points.

B' : symmetric ot B with respect to A, M middle of BC, D : trace on BC of bisector of A.

E : trace on CB' of external bisector of A.

By the basic bisector relation ( Bisector0.html ).

DB/DC = AB/AC = AB'/AC = EB'/EC, hence DE is parallel to BB'.

It follows that the right-angled triangle DAE is constructible since |AD|=b

i) the angle at A, ii) the bisector |AE| and iii) the side |CB'| = 2m

This kind of construction is discussed in PappusTriangleConstruction.html .

Bisector0.html

Bisector1.html

BisectorRectangle.html

Euler.html

PappusTriangleConstruction.html

TriaConstructionAarb.html

TriangleBisectors.html

Fursenko F. B.

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