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EXERCISES.

1. Show that the straight line which bisects the external vertical angle of an isosceles triangle is parallel to the base.

2. A straight line is drawn terminated by two parallel straight lines; through its middle point any straight line is drawn and terminated by the parallel straight lines. Show that the second straight line is bisected at the middle point of the first.

3. Show that the angle between the bisector of the angle A of the triangle A B C and the perpendicular let fall from A on BC is equal to one-half the difference between the angles B and C.

4. In any right triangle show that the straight line drawn from the vertex of the right angle to the middle of the hypotenuse is equal to one-half the hypotenuse.

5. Two tangents are drawn to a circle at opposite extremities of a diameter, and cut off from a third tangent a portion A B. If C be the centre of the circle, show that A C B is a right angle.

6. Show that the sum of the three perpendiculars from any point within an equilateral triangle to the sides is equal to the altitude of the triangle.

7. Show that the least chord which can be drawn through a given point within a circle is perpendicular to the diameter drawn through the point.

8. Show that the angle contained by two tangents at the extremities of a chord is tvice the angle contained by the chord and the diameter drawn from either extremity of the chord.

9. If a circle can be inscribed in a quadrilateral ; show that the sum of two opposite sides of the quadrilateral is equal to the sum of the other two sides.

10. If the sum of two opposite sides of a quadrilateral be equal to the sum of the other two sides; show that a circle can be inscribed in the quadrilateral.

ON PROPORTIONAL LINES.

PROPOSITION I. THEOREM. 274. If a series of parallels intersecting any two straight lines intercept equal parts on one of these lines, they will intercept equal parts on the other also.

H

K
Let the series of parallels A A', B B', C C', D D', E E',

intercept on H' K' equal parts A' B', B'C', C'D', etc.
We are to prove
they intercept on H K equal parts A B, BC, C D, etc.
At points A and B draw A m and Bn || to H' K'.
A m = A' B'.

§ 135
(parallels comprehended between parallels are equal).
Bn= B'C',

$ 135 .. Am= B n. In the A B Am and C Bn, ZA= LB,

§ 77 (having their sides respectively || and lying in the same direction from

the vertices).
Z m = Zn,

§ 77 and

Am= Bn,
.. A B Am = AC Bn,

$ 107 (having a side and two adj. & of the one equal respectively to a side and

two adj. e of the other).

.. AB=BC,

(being homologous sides of equal A).
In like manner we may prove BC= C D, etc.

Q. E. D.

PROPOSITION II. THEOREM. 275. If a line be drawn through two sides of a triangle parallel to the third side, it divides those sides proportionally.

Fig. 1.

Fig. 2.
In the triangle A B C let E F be drawn parallel to BC.

EB FC
We are to prove

A E AF
CASE I. — When A E and EB (Fig. 1) are commensurable.
Find a common measure of A E and E B, namely B m.
Suppose B m to be contained in B E three times,

and in A E five times.
Then

EB _ 3

A E 5 At the several points of division on B E and A E draw straight lines II to BC.

These lines will divide A C into eight equal parts, of which FC will contain three, and A F will contain five, § 274 (if parallels intersecting any two straight lines intercept equal parts on one of these lines, they will intercept equal parts on the other also).

. FC 3

AF - 5
But

EB 3

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Ax. 1

CASE. II. When A E and E B (Fig. 2) are incommensurable.

Divide A E into any number of equal parts, and apply one of these parts to E B as often as it will be contained in E B.

Since A E and E B are incommensurable, a certain number of these parts will extend from E to a point K, leaving a remainder K B, less than one of the parts.

Draw K H || to BC.
Since A E and E K are commensurable,
EK FH

(Case I.) A E AF Suppose the number of parts into which A E is divided to be continually increased, the length of each part will become less and less, and the point K will approach nearer and nearer to B. The limit of E K will be E B, and the limit of FH will be FC.

will be

.. the limit of Ek will be EB

A E'
FC

and

the limit of **, will be

[ocr errors]
[ocr errors]

Now the variables

and FH

4. are always equal, how

" Α Ε Α F' ever near they approach their limits;

Е В , FC is their limits

are equal,

§ 199

Q. E. D. 276. COROLLARY. One side of a triangle is to either part cut off by a straight line parallel to the base, as the other side is to the corresponding part. Now EB : AE :: FC : AF.

§ 275 By composition,

EB + A E : AE :: FC + AF : AF, § 263 or,

A B : A E :: AC : A F.

PROPOSITION III. THEOREM. 277. If a straight line divide two sides of a triangle proportionally, it is parallel to the third side.

In the triangle A B C let E F be drawn so that

of A B AC

hat AE AF We are to prove EF || to B C.

From E draw E H || to B C.
Then
AB AC

$ 276 A E - AH' (one side of a A is to either part cut off by a line Il to the base, as the other

side is to the corresponding part).
But

AB AC
A E = AF

Hyp.
..AC AC
ĀF = AH

Ax. 1 .. AF = A H.

.:. EF and E H coincide,

(their extremities being the same points).
But
E H is II to BC;

Cons. .. E F, which coincides with E H, is II to B C.

Q. E. D. 278. DEF. Similar Polygons are polygons which have their homologous angles equal and their homologous sides proportional.

Homologous points, lines, and angles, in similar polygons, are points, lines, and angles similarly situated.

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