« ΠροηγούμενηΣυνέχεια »
PROPOSITION XXXIV. THEOREM.
118. Of two angles of a triangle, that is the greater which is opposite the greater side.
In the triangle ABC let A B be greater than A C.
Take A E equal to AC;
§ 105 (an exterior 2 of a A is greater than either opposite interior 2),
ZACB > LACE.
Q. E. D.
Ex. If the angles A B C and AC B, at the base of an isosceles triangle, be bisected by the straight lines BD, CD, show that DBC will be an isosceles triangle.
PROPOSITION XXXV. THEOREM. 119. The three bisectors of the three angles of a triangle meet in a point.
Let the two bisectors of the angles A and C meet
at 0, and 0 B be drawn.
Draw the Is OK, O P, and 0 H.
$ 110 (having the hypotenuse and an acute 2 of the one equal respectively to the
hypotenuse and an acute of the other).
(homologous sides of equal A).
Cons. .. AO AP=A O A H,
§ 110 (having the hypotenuse and an acute L of the one equal respectively to the
hypotenuse and an acute Z of the other).
OH = 0 K, and 0 B = 0 B,
$ 109 (having the hypotenuse and a side of the one equal respectively to the hypote
nuse and a side of the other),
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.
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.
Then, since O is in D D', which is I to A B at its middle point, it is equally distant from A and B.
§ 59 Also, since () is in E E', I to A C at its middle point, it is equally distant from A and C.
..O is equally distant from B and C;
..O is in FF 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.
In the triangle ABC, let B P, A H, CK, be the per
pendiculars from the vertices to the opposite
A' B' || to BC,
B'C' || to A B.
§ 68 (being alternate interior ), Z BA A' = Z A B C.
§ 68 .. A A BA' = A A BC,
§ 107 (having a side and two adj. s of the one equal respectively to a side and
two adj. e of the other).
.. A' B = A C,
In the ACBC and A B C,
Iden. Z C BC= _ BCA,
§ 68 (being alternate interior &). < BCC' = Z CBA.
§ 68 .. ACBC" = A ABC,
§ 107 (having a side and two adj. of the one equal respectively to a side and two
adj. As of the other).
..BC = AC,
.. B is the middle point of A' C'.
Since B P is I to A C,
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';
.:. B Pis 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 K I 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.
$ 120 (the three Is erected at the middle points of the sides of a A meet in a point).
Q. E. D.