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PROPOSITION XXIX. THEOREM.

113. A straight line which bisects the angle at the vertex af an isosceles triangle divides the triangle into two equal triangles, is perpendicular to the base, and bisects the base.

Let the line C E bisect the Z A C B of the isosceles

A ACB.

We are to prove I. A ACE= A BCE;

II. line CEI to A B;

III. A E= B E. 1. In the A AC E and BCE, AC=BC,

Hyp. CE=CE,

Iden. LACE= Z BCE.

Cons. .. A ACE= A BCE,

§ 106 (having two sides and the included L of the one equal respectively to two sides

and the included L of the other). Also, II.

ZCEA=LCEB,

ZCEA =
(being homologous 5 of equal A).

..CE is I to A B, (a straight line meeting another, making the adjacent & equal, is I to

that line).
Also, III.

A E= EB,
(being homologous sides of equal A).

Q. E. D.

PROPOSITION XXX. THEOREM.

114. If two angles of a triangle be equal, the sides opposite the equal angles are equal, and the triangle is isosceles.

BL

In the triangle A BC, let the Z B= 2 C.
We are to prove A B = AC.

Draw A D I to BC.
In the rt. A A D B and A DC,
AD= AD,

Iden.
ZB=2C,

.:. rt. A À D B=rt. A A DC, $ 111 (having a side and an acute. Z of the one equal respectively to a side and an

acute 2 of the other).

.. AB= AC,
(being homologous sides of equal A).

Q. E. D.

- Ex. Show that an equiangular triangle is also equilateral.

Proposition XXXI. THEOREM. 115. If two triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first will be greater than the third side of the second.

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Hyp:

In the A ABC and A B E, let A B = A B, BC=BE; but 2 ABC > Z A B E.

We are to prove AC > A E. Place the A so that A B of the one shall coincide with A B of the other.

Draw B F so as to bisect Z EBC.

Draw EF.
In the A EBF and CBF

EB=BC,
BF=BF,

Iden.
ZEBF= LCBF,

Cons. .. the A E B F and C B F are equal, $ 106 (having two sides and the included 2 of one equal respectively to two sides

and the included 2 of the other).

.. EF= FC,

(being homologous sides of equal A).
Now
A F + FE > A E,

§ 96 (the sum of two sides of a A is greater than the third side). Substitute for F E its equal FC. Then

AF + FC > A E; or,
AC > A E.

Q. E. D.

PROPOSITION XXXII. THEOREM.

116. CONVERSELY: If two sides of a triangle be equal respectively to two sides of another, but the third side of the first triangle be greater than the third side of the second, then the angle opposite the third side of the first triangle is greater than the angle opposite the third side of the second.

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§ 106

In the A ABC and A' B'C', let A B = A' B', AC = A' C'; but BC > B'C'. .

We are to prove ZA > LA'.
If

ZA = L A', then would A ABC= A A' B'C', (having two sides and the included 2 of the one cqual respectively to two sides

and the included 2 of the other), and

BC= B'C',

(being homologous sides of equal A ). And if

A <A', then would BC < B'C',

§ 115 (if two sides of a A be equal respectively to two sides of another A, but the

included L of the first be greater than the included Z of the second, the third side of the first will be greater than the third side of the second.)

But both these conclusions are contrary to the hypothesis ; ..Z A does not equal Z A', and is not less than 2 A'.

..LA>A'.

Q. E. D

PROPOSITION XXXIII. THEOREM.

117. Of two sides of a triangle, that is greater which is opposite the greater angle.

In the triangle A B C let angle ACB be greater than

angle B.

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Draw C E so as to make Z BCE= Z B.

Then
EC= EB,

$ 112
(being sides opposite equal £ ).
Now
A E + EC > AC,

§ 96 (the sum of two sides of a A is greater than the third side). Substitute for EC its equal E B. Then

A E + EB > A C, or
A B > A C.

Q. E. D.

Ex. A B C and A B D are two triangles on the same base A B, and on the same side of it, the vertex of each triangle being without the other. If A C equal A D, show that BC cannot equal B D.

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