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PROPOSITION XXXIV. THEOREM.

118. Of two angles of a triangle, that is the greater which is opposite the greater side.

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In the triangle A B C let A B be greater than AC.

We are to prove

LACB> LB.
Take A E equal to AC;

Draw EC.

§ 112

LAEC = LACE,

(being & opposite equal sides). But LAEC > LB,

$ 105 (an exterior 2 of a A is greater than either opposite interior 2),

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Ex. If the angles ABC and ACB, at the base of an isosceles triangle, be bisected by the straight lines BD, CD, show that D B C will be an isosceles triangle.

PROPOSITION XXXV.

THEOREM. 119. The three bisectors of the three angles of a triangle meet in a point.

B

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Let the two bisectors of the angles A and C meet

at 0, and O B be drawn.

OK,

We are to prove

BO bisects the < B.

Draw the Is OK, O P, and O H.
In the rt. A OC K and OCP,
0 C =0 C,

Iden.
LOCK= LOCP,

Cons.
AOCK=AOCP,

$ 110 (having the hypotenuse and an acute < of the one equal respectively to the

hypotenuse and an acute 2 of the other).

..OP

(homologous sides of equal ).
In the rt. A 0 A P and 0 A H,
OA=0 A,

Iden.
LOAP=LOAH,

Cons.
..ΔΟ ΑΡ.
ΔΟ Α Η, ,

§ 110 (having the hypotenuse and an acute 2 of the one equal respectively to the

hypotenuse and an acute Z of the other).

.. OP:
(being homologous sides of equal a ).
But we have already shown 0 P=0 K,
.:.011= OK,

Ax. 1
Now in rt. A OH B and O KB

OH,

1

OH= 0 K, and OB= 0 B,
...ΔΟ Η Β =ΔΟ Κ Β,

$ 109 (having the hypotenuse and a side of the one equal respectively to the hypote

nuse and a side of the other),
..LOBH= LOBK,
(being homologous & of equal A).

Q. E. D.

PROPOSITION XXXVI. - THEOREM. 120. The three perpendiculars erected at the middle points of the three sides of a triangle meet in a point.

A

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Let D D, E E, F F, be three perpendiculars erected

at D, E, F, the middle points of A B, A C, and BC.

We are to prove they meet in some point, as 0.

The two Is D D' and E E' meet, otherwise they would be parallel, and A B and A C, being is to these lines from the same point A, would be in the same straight line;

but this is impossible, since they are sides of a A.
Let O be the point at which they meet.

Then, since 0 is in D D', which is I to A B at its middle point, it is equally distant from A and B.

Also, since O is in E E', I to A C at its middle point, it is equally distant from A and C.

$ 59

..O is equally distant from B and C;

:: 0 is in F F' I to B C at its middle point, $ 59 (the locus of all points equally distant from the extremities of a straight line

is the I erected at the middle of that line).

Q. E. D.

PROPOSITION XXXVII. THEOREM.

121. The three perpendiculars from the vertices of a triangle to the opposite sides meet in a point.

B!

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In the triangle A BC, let B P, A H, CK, be the per

pendiculars from the vertices to the opposite sides.

We are to prove they meet in some point, as 0.

[blocks in formation]

..Δ Α Β Α'
= A A BC,

§ 107 (having a side and two adj. s of the one equal respectively to a side and

two adj. s of the other).

.:. A B
(being homologous sides of egual D ).

AC,

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.:. ДС В С. A ABC,

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

adj. e of the other).

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it is I to A' C',

$ 67 (a straight line which is I to one of two lls is I to the other also).

But B is the middle point of A'C';

.. BP is I to A' C' at its middle point.

In like manner we may prove that

A H is I to A' B' at its middle point,

and C KI to B'C' at its middle point.

.. BP, A H, and C K are Is erected at the middle points of the sides of the A A' B'C'.

::. these Is meet in a point. (the three Is erected at the middle points of the sides of a A meet in a point).

§ 120

Q. E. D.

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