PROPOSITION 39. THEOREM.

Equal triangles on the same base and on the same side of it are between the same parallels.

Let ABC and DBC be equal ▲s on the same base BC, and on the same side of it, they shall be between the same parallels.

A

B

D

Join AD.

AD shall be || to BC.

For if it is not, through A draw AE || to BC, meeting BD in E,

In like manner it can be shown that no other straight line through A but AD is to BC.

.. AD is || to BC.

(For it has been shown in Prop. 31 that there can always be found one parallel through A to BC.)

NOTES.

Ex. 81. To bisect a quadrilateral by a line drawn through a given vertex. If the diagonal DA through the given point ▷ divides the given quadrilateral into two unequal triangles, through B, the vertex of the smaller one, draw a line BE, making a triangle, DEC, equivalent to the given quadrilateral, and draw its median DF.

Ex. 82. To bisect a triangle by a line drawn through a given point in one of its sides. If the line DA joining the given point ↳ to the opposite vertex A divides the given triangle into two unequal triangles, through B, the vertex of the smaller one, draw a line BE, making a triangle, DEC, equivalent to the given triangle, and draw its median DF.

(Note the close agreement between this and the previous solution: we treat ABDC as if it were a quadrilateral with D for one vertex.)

From Corollary to I. 37 and I. 39: 'The locus of the vertex of a triangle on a given base, and having a given area, is a pair of straight lines parallel to the given base, and on opposite sides of it.'

Ex. 83.-ABC and DEF are any two triangles. Find a point P such that the two triangles PBC, PEF are equal respectively to ABC, DEF. P lies on two different loci, and will be at their intersection if there be such a point. Show that if there is such a point there must at least be three others possessing the same property, and that there may be an infinite number.

Ex. 84.—If two equivalent triangles ABC, DBC be on the same base BC, and on opposite sides of it, the straight line AD will be bisected by BC or BC produced.

Compare Ex. 68, of which this is a converse.

Equal triangles on equal bases in the same straight line and on the same side of it are between the same parallels.

Let ABC and DEF be equal As on equal bases BC and EF, in the same straight line BF, and on the same side of it, they shall be between the same parallels.

Join AD. AD shall be || to BF.

E

For if it is not, through A draw AG || to BF, meeting ED in G,

In like manner it can be shown that no other straight line

through A but AD is || to BF.

.. AD is to BF.

Ex. 85.-Enunciate, and prove another converse of I. 38.

if

PROPOSITION 41. THEOREM.

a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram shall be double of the triangle.

Let ||gm ABCD and ▲ EBC be on the same base BC, and between the same parallels BC and AE, then ||gm ABCD shall be double of ▲ EBC.

A

E

B

Join AC.

Then ABC = AEBC.

But ||gm ABCD is double of the ▲ ABC.

... the ||gm ABCD is double of the

[I. 37.

[I. 34.

EBC.

Ex. 86. If two equivalent triangles ABC, DEF be on equal bases BC, EF, in the same straight line BF, and on opposite sides of it, the straight line AD will be bisected by BF.

Show that the

Ex. 87.-A fixed straight line AD is bisected by any other straight line BF. On BF are taken any equal segments BC, EF. triangles ABC, DEF are equivalent.

Ex. 88.-Find a set of lines such that any segment of any one of them is the base of two equivalent triangles, having two given points as vertices. The set consists of two bencils, one of parallels and the other of con. current lines.