Divifion of Right Lines , Surfaces , and Solids. 87 
make equal angles with the hypotenufe , < 2 /^ the right line 
drawn from the right angle to meet it; and will likewife 
have to each other the proportion of the fides containing 
the right angle (<). 
COR. 
X *• • • 
(c) For in the triangle acb, having the angle at c right, conftru£l the figure 
as in the firfi: propofition, and join lp, Lq^ The author affirms, that the angles 
13 
pl a, qlb, and alfo the angles clp, qlc, are equal, and that pl : lqjtca : cb. 
Produce lp, lq^, till they meet the right line ih in m, n. Now, becaufe mn, ab, 
are parallel, therefore al : mc=ap : pc, and cn : LB — c<^ (or pc) : qB. 
Therefore, al x cn : mc x lb=tap : qb. But af : qb — al : lb (by theor. i. 
cor. 5.). Therefore, al x cn : mc x lb rr al : lb. Therefore, cn and cm arc 
equal. But becaufe the angle acb is right, cl is perpendicular to ab (by 
theor. 1. cor. 2.), and confequently to mn. Therefore the angles mcl, ncl, 
are right, and mc being equal to cn, and cl common to the two triangles mcl, 
ncl, ml will be equal to ln, and the angles mlc, cml, equal to nlc, CNL,refpec- 
tively. Hence it is evident, that the angles clp, clq^, are equal, and each is 
half a right angle ; and likewife pla, qlb, are equal, and each half a right angle. 
F urther. 
