same thing also wil happen if the side BA be deuided into two equall parts in the point E, and so be drawen a right line from the point E, to the point C. Or if the side AC be deuided into two equall partes in the point F, and so be drawen a right line from the point F to the point B: which is in like manner proued by drawing parallel lines by the pointes B, and C, to the lines BA and AC,
And so by this you may deuide any triangle into so many partes as are sig∣nified by any number that is euenly euen: as into 14.16.32.64. &c.
An other addition of Pelitarius.
From any point geuen in one of the sides of a triangle, to draw a line which shal deuide the trian∣gle into two equall partes.
Let the triangle geuen be BCD: and let the point geuen in the side BC be A. It is required from the point A to draw a line which shal deuide the triangle BCD into two equall partes. Deuide the side BC into two equall partes in the point E. And drawe a right line from the point A to the point D. And (by the
31. proposition) by the point
E draw vnto the line
AD a parallel line
EF: which let cutte the side
DC in the point
F. And draw a line from the point
A to the point
F. Then I say that the line
AF deuideth the triangle
BCD into two equall partes: namely, the trapesium
ABDF is equall to the triangle
ACF. For draw a line from
E to
D, cutting the line
AF in the point
G. Now then it is manifest (by the 38. proposition) that the two trian∣gles
BED and
CED are equall (if we vnderstand a line to be drawen by the point D parallel to the line
BC, for the bases
BE and
EC are equal). The two triangles also D
EF and
AEF are (by the 37. proposition) equall: for they consist vpon one and the selfe same base
EF, and are in the selfe same parallel lines
AD and
EF. Wherefore taking away the triangle
EFG which is cōmō to thē both, the triangle
AEG shalbe equall to the triangle D
FG: wher¦fore vnto either of thē adde the trapesiū
CFGE, and the triangle
ACF shalbe equal to the triangle
DEC. But the triangle
DEC is the halfe part of the whole triangle
BCD wherefore the triangle
ACF is the halfe part of the same triangle
BCD. Wherfore the residue, namely, the trapesium
ABFD is the other halfe of the same triangle. Where∣fore the line
AF deuideth the whole triangle
BCD into two equall partes: which was required to be done.