1833. | Trisection of an Angle. 71 
III.—Trisection of an Angle. By Col. Nasmyth Morrieson. 
Proposition 1st, Theorem. 
If, from the vertical angle of a triangle, having one of the angles at 
its base double of the other one, and the vertical angle greater than 
half a right angle, a straight line be drawn to cut the base, making an 
angle with the greater side of the triangle adjacent to the vertical 
angle, equal to the lesser angle at the base; and if from the vertical 
angle as a centre, at the distance of the lesser side of the triangle ad- 
jacent to the vertical angle, a circle be described ; the circle, and the 
' line drawn from the vertical angle to cut the base, and the base of the 
triangle, have one common intersection. 
Let ABC (fig. 1) be a triangle, having the angle BAC, one of the angles 
at its base double of the angle BCA, the other angle at its base, and its 
vertical angle ABC greater than halfa right angle; and let the straight 
line BD be drawn from the vertical angle ABC, to cut the base ACG in 
D, making with CB, the greater side of the triangle adjacent to the 
vertical angle, the angle CBD equal to BCA, the lesser angle at the 
base (23.1) ; also from B as acentre at the distance BA, the lesser side 
of the triangle adjacent to the vertical angle, let the circle ADE be 
described; the circle ADE, the line BD, and the base AC intersect one 
another in one common point D. 
Because, by construction, the angle DBC is equal to the angle DCB, 
the side BD is equal to the side CD (5.1), and D is the point of inter- 
section of BD and AC. Again, because BDA, the exterior angle of the 
triangle BDC, is equal to the two interior and opposite (82.1) and also 
equal angles DBC, DCB, therefore, BDA is double of DCB, that is ACB; 
but, by construction, the angle BAC is double of ACB, therefore BAC 
is equal to BDA (6 ax); and because the angle BAC is equal to the 
angle BDA, the side BD is equal to the side BA (5.1); wherefore 
the circle ADE described from the centre B, at the distance BA passes 
through D, the extremity of BD, or D is the point of intersection 
of the circle ADE and the line BD; but it bas been already shewn 
that D is the point of intersection of BD and AC, consequently the 
circle ADE intersects, in the point D, the line AC ; therefore the circle 
ADE, and the straight lines AC and BD intersect in one common point 
D: ».Q.E: D. 
Proposition 2nd, Problem. 
To draw the base of a triangle, so that, of the interior angles at the 
base, one shall be double of the other, the vertical angle of the triangle 
being a given rectilineal angle greater than half a right angle. 
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