Suppose that there be two right lines AB and GD cutting the one the other in the point E, and making an obtuse angle in the section E. And from the endes of the lines, namely, A and G, let there be drawen to either line perpendicular lines, namely, from the point A to the line GD, which let be AD, and from the point G to the right line AB: which let be GB. Then I say, that the
right lines
AB and
GD do, betwene the end
A and the perpendicular
B, and the end
G and the perpendicular
D, cut the one the other reciprokally in the point
E: so that as the line
AE is to the line
ED, so is the line
GE to the line
EB. For foras∣much as the angles
ADE and
GBE, are right angles, therfore they are equall. But the angles
AED and
GEB are also e∣quall (by the 15. of the first). Wherefore the angles remayning, namely,
EAD, &
EGB, are equall (by the Corollary of the 32. of the first). Wherefore the triangles
AED and
GEH, are equiangle. Wherfore the sides about the equall angles shall be proportionall (by the 4. of the sixt). Wherfore as the line
AE is to the line
ED, so is the line
GE to the line
EB. If therefore two right lines cut the one the other obtuseangled wife. &c: which was required to be proued.
¶ The third Proposition.
If two right lines make an acute angle, and from their endes be drawen to ech line perpendicular lines cutting them: the two right lines geuen shall be reciprokally pro∣portionall as the segmentes which are about the angle.
Suppose that there be two right lines AB and GB, making an acute angle ABG. And from the poyntes A and G let there be drawen vnto the lines AB and GB per∣pendicular lines AC and GE, cutting the lines AB and GB in the poyntes E and ••. Then I say, that the lines, namely AB to GB, are reciprokally proportionall, as the seg∣mentes, namely, CB to EB which are about the acute
angle
B. For forasmuch as th
•• right angles
ACB and
GER are equall, and the angle
•• ABG is common to the triangles
ABC, and
GBE•• therefore the angles re∣mayning
BAC and
EGB are equall (by the Corolla∣ry of the 32. of the first). Wherfore the triangles
ABC and
GBE are equiangle. Wherefore the side
•• about the equall angles are proportionall (by the 4. of the sixe)
•• so that, as the line
AB is to the line
FC, so is the line
GB to the line
BE. Wherefore alternately as the line
AB is to the line
GB so is the line
CB to the line
BE. If therefore two right lines mak
•• a
•• ••c
••te angle
•• &c
•• which was required to be proued.
•• The fourth Proposition••
If in a circle be drawen two right lines cutting the one the other, the sections of the one to the sections of the other shall be reciprokally proportionall.
In the circle, AGB let these two right lines 〈…〉〈…〉 one the other in the poynt E. Th•••• I say, that reciprokally 〈◊〉〈◊〉 ••h•• line AE is to the line ED, so is the line GE to the line EB. For forasmuch as (by the 35. of the third) the rectangle fi∣gure