b = AH. and join FC, and IC; and produce ACG.
c conſlr.
d 15.1.
e 4.1.
f 15.1.
g 9.ax.Becauſe CE c=EA, and EF c=EB, and the angle FEC d = BEA, the angle ECF e ſhall be equal to EAB. By the like argument is the angle ICH = ABH. Therefore the whole angle ACD (f BCG) g is greater than either the angle CAB or ABC. Which was to be demonſtrated.
PROP. XVII.
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Two angles of any triangle ABC, which way ſoever they be taken, are leſs than two right angles.
Let the ſide BC be produced. Becauſe the angle ACD + ACB a = 2 right a 13.1.
b 16.1.
c 4.ax.angles, and the angle ACD b 
A, c therefore A + ACB 
then two right angles. After the ſame manner is the angle B + ACB then two right. Laſtly, the ſide AB being produced, the angle A + B will be alſo leſs than two right angles. Which was to he demonſtrated.
Coroll.
1. Hence it follows that in every triangle, wherein one angle is either right or obtuſe, the two others are acute angles.
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2. If a right line AE make unequal angles with another right line D, one acute AED, the other obtuſe AEC, a perpendicular AD let fall from any point A to the other line CD, ſhall fall on that fide the acute is of.
For if AC, drawn on the ſide of the obtuſe angle, be a perpendicular, then in the triangle AEC * 17.1. ſhall AEC + ACE be * greater than two right angles. Which is contrary to the prectdent Prop.
3. All the angles of an equilateral triangle, and the two angles of an Iſoſceles triangle that are upon the baſe, are acute. PROP.
