PROPOSITION XXXIV. THEOREM.

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

E

B

Α

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

(an exterior of a ▲ is greater than either opposite interior ▲),

and

ZACBZACE.

Substitute for ACE its equal AEC, then

ZACB ZAEC.

Much more is Z ACB > Z B.

§ 112

§ 105

Q. E. D.

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.

119. The three bisectors of the three angles of a triangle

Let the two bisectors of the angles A and C meet at 0, and OB be drawn.

Draw the Is OK, O P, and O H.

In the rt. A OCK and OCP,

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

hypotenuse and an acute of the other).

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

OH = 0 K, and 0 B = 0 B,

.. ΔΟΗΒ ΔΟΚΒ,

$ 109

(having the hypotenuse and a side of the one equal respectively to the hypote nuse and a side of the other),

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

Let D D', EE, FF', be three perpendiculars erected at D, E, F, the middle points of A B, A C, and B C.

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

The two DD' 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 ▲.

Let O be the point at which they meet.

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

$ 59

Also, since is in EE', 1 to AC at its middle point, it is equally distant from A and C.

.. O is equally distant from B and C ;

.. O is in FFL to BC at its middle point,

$59

(the locus of all points equally distant from the extremities of a straight line is the 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.

In the triangle ABC, let BP, AH, CK, be the perpendiculars from the vertices to the opposite sides.

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

Through the vertices A, B, C, draw

(having a side and two adj.

of the one equal respectively to a side and two adj. of the other).

(having a side and two adj. 4 of the one equal respectively to a side and two

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

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

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

In like manner we may prove that

AH is to A' B' at its middle point,

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

Hyp.

$ 67

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

.. these Is meet in a point.

$120

(the three s erected at the middle points of the sides of a ▲ meet in a point).

Q. E. D.