PROPOSITION XXIII. PROBLEM

117. At a point C, in a given straight line AB, to construct an angle equal to a given angle O.

A

Да

Given. Point Cin line AB and ≤ 0.

Construction. From O as a center, with any radius, describe an arc cutting the sides of 20 in D and E.

From C as a center, with the same radius, describe arc FG, intersecting CB in F. From F as a center, with a radius equal

to DE, draw an arc intersecting FG in H.

Join CH.

LHCF is an at C = ≤0.

HINT. What is the usual means of proving the equality of angles?

NOTE.

Q.E.F.

- Two letters, e.g. DE, used as above, denote a straight line.

PROPOSITION XXIV. PROBLEM

118. Through a given point to draw a par

Construction. Through O draw DOC, forming ≤D w At O construct COB = ODB.

A'B' is a line through O || AB.

What is the usual means of proving that two lines are

Ex. 159. From a given point without a line to draw a line, given angle with the given line.

Ex. 160. Through a given point to draw a parallel to a give means of (a) Prop. V, (b) Ex. 50.

PROPOSITION XXV. THEOREM

119. Every point in the perpendicular-bisecto line is equidistant from the extremities of tha

D

Hyp. PD is the perpendicular-bisector of AB.

C is a point in PD.

To prove C is equidistant from A and B.

What is the usual means of proving the equality of lines?

Ex. 161. In a given line AB find a point equidistant from two given points P and Q.

Ex. 162. In a given circumference find a point equidistant from two given points.

Ex. 163. Find a point equidistant from three given points.

120. DEF. The distance of a point from a line is the length of the perpendicular from the point to the line.

PROPOSITION XXVI. THEOREM

121. All points in the bisector of an angle are equidistant from the sides of the angle.

Hyp.

DAB = Z DAC and O is any point in AD.

To prove O is equidistant from AB and AC.

Draw OP LAB and OP' L AC, and prove the equality of the two triangles.

122. COR. A point equidistant from the sides of an angle is in the bisector of that angle.

UNEQUAL LINES AND UNEQUAL ANGLES

PROPOSITION XXVII. THEOREM

123. The sum of two sides of a triangle is and their difference is less, than the third si

Hyp.

To prove

B

ACB is a A.

AB+ AC > BC; and BA> CB – AC

Proof. 1°. A straight line is the shortest distance

Ex. 167. Prove that the perimeter of poly- B
gon ABCDEF> perimeter of ▲ ACE.

Ex. 168. If from any point D in AABC
DB and DC are drawn, then

AB+ AC>DB + DC.

PROPOSITION XXVIII. THEOREM

124. If two sides of a triangle are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side.

Proof. Draw AE, bisecting BAC. On AC lay off AD AB and draw DE.

(exterior of a ▲ is > either remote interior ≤).