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Commenced Oetitoon hionday
On a given finite right line to construct an equilateral
triangle. With the extremities of the given line as centres and the given line as radius describe two circles, (per post.3.); from a point in which those circles intersect one another draw lines to the extremities of the given one; those three lines form an equilateral triangle.
For each of the drawn lines is equal to the given line as being radii of the same circle, and they are ::= to each other (per ax. 1.)
Note.- No more than two equilateral triangles can be formed on the same right line, viz. one on each side of it.
Those circles willintersect; because, the centre of either being in the circumference of the other, part of its circumference must be within the other.
PROP. 2, PROB.
From a given point to daw a right line equal to a given
finite right line. Connect the given point with either extremity of the given line; on this connecting line form an equilat. trian.
(Prop. 1); with the connected extremity of the given line as centre, and the given line as radius, describe a circle whose circumference shall meet the leg of the equilateral triangle produced, through the connected point; with the vertex of equilat. tri. as centre, and whole produced leg, as radius, describe another circle to meet the other leg produced through the given point, this produced part is the line required to be drawạ.
For it is = to the produced part of the other leg ; since the whole produced legs are =, and the parts of them which are the sides of the equilat. tri. are also = .. the remainders, viz. the produced parts are = but the part produced from the connected point is = to the given line as being radii of the same circle, ., the part produced from the given point is = to the given line.
PROP. 3, PROB.
To cut from the greater of two given right lines a part
equal to the less. From either extremity of the greater draw a line = to the less (per Prop. 2,) and with the same extremity as a centre and the drawn line as an interval describe a circle; this circle cuts from the greater line, a part,= to the less.
For the part cut off between the circumference and centre js = to the drawn line, which was made to the lesser of the given lines in this part is = to the lesser of the given lines.
PROP. 4, THEOR. If two triangles have two sides of the one respectively
equal to two sides of the other, and the angle contained by those two sides of the one equal to the angle contained by the two sides equal to them of the other, their bases or third sides shall be equal, also the angles at the bases shall be equal, viz. those opposite the equal sides, and the whole triangles shall be equal.
If the two triangles be so applied to one another, that the vertex of one may coincide with that of the other, and a side of one coincide with a side equal to it of the other, and the other two sides lie towards the same part. Then since the angles at the vertices are = and the vertices coinciding the sides about those angles must coincide, and since they coincide, and are =, their other extremities must coincide, and since those extremities coincide, (which are at the bases,) the bases themselves must coincide (per ax. 10.), and be
And since the bases (and other sides) coincide the angles at the bases must be =, and the sides and anglès coinciding and being respectively =, the whole triangles must coincide and be =.
PROP. 5, THEOR.
In an isosceles triangle the angles at the base are equal
to one another; and if the equal sides be produced the angles below the base shall be also equal.
Assume points on the legs below the base equidistant from the vertex; and from these assumed points draw transverse lines to the opposite extremities of the base ; then there are two triangles formed, having a side of isosceles triangle and cut off seg. in one, respectively = to a side of isosceles triangle and cut off seg in the other, (by the 4th,)the bases, (viz. the trans. lines) are, and also the angles at the bases, (viz. those contained by sides of isos..tri. and trans. lines, and those contained by cut off seg. and trans. lines) are respectively; and in the triangles below the base the angles opposite the base are = (per demon.) and the sides containing them are respectively ::(by 4th) the angles below the base are = and also the angles contained by the base and tranverse lines, which being taken from the angles contained by the sides of isos. tri. and trąnverse lines leave the angles at the base = (per ax 3.)
Hence every equilateral triangle is also equiangular. Vide Elrington,
PROP: 6. THEOR.
If two angles of a triangle be equal the sides opposite to
them are also equal, For if they be pot = let one of them be greater, and from it cut off a part (towards one of the = angles) to the lesser ; and draw a trap verse line from the point of section to the opposite angle; theu there are two triangles, formed having two sides in each respectively and the
contained angles also = (by hypoth), the triangles are =, a part to the whole which is absurd.
Cor. Hence every equiangular triangle is equilateral; vide, Elrington.
PROP. 7. THEOR.
On the same right line and on the same side of it there
cannot be two triangles formed whose conterminous sides are equal. If it be possible that there can, Ist, let the vertex of one fall without the other, draw a right connecting their vertices, then this connecting line will be the base of two isosceles triangles, and the angles at the base must be = in each, but if they be = in one of these triangles, in the other, one of them must be much greater than the other, and also = to it which is absurd, (one of them being greater and the other less than one of the former pair.)
2ndly, If the vertex of one triangle be supposed to fall within the other, connect their vertices and produce either pair of = legs. Then, the angles below the base of one isosceles triangle are = (by the 5th ;) and of the angles above the base of the other isosceles triangle, one of them is much greater than the other, but they are also =, which is absurd.
3dly, If the vertex of one triangle be supposed to fall upon a side of the other, it is evident that two of the conterminous sides are not equal.
PROP. 8, THEOR,
If two triangles have two sides of the one respectively,
equal to troo sides of the other, and also have the base of the one equal to the base of the other, the angles contained by the equal sides are also respectively equal.
For if the = bases be so applied to one another, that the sides respectively = may be conterminous and that the triangles lie towards the same parts, the vertices must fall
upon one another, (per prop. 7,) and the = sides must coincide, (per ax. 10,).". the angles must coincide, and