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PROPOSITION XXIV. THEOREM. 107. Two triangles are equal in all respects when a side and two adjacent angles of the one are equal respectively to a side and two adjacent angles of the other.

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In the triangles A B C and A' B' C', let A B = A' B',

ZA= 2 A', Z B= 2 B'.

We are to prove A ABC= A A' B'C'.

Take up A ABC and place it upon A A' B'C', so that A B shall coincide with A' B'.

AC will take the direction of A' C',

(for ZA= L A', by hyp.); the point C, the extremity of A C, will fall upon A' C' or A' C' produced

BC will take the direction of B'C',

(for ZB= 2 B', by hyp.) ; the point C, the extremity of BC, will fall upon B'C' or B'C' produced.

.. the point C, falling upon both the lines A' C' and B' C', must fall upon a point common to the two lines, namely, C'.

.. the two o coincide, and are equal in all respects.

Q. E. D.

PROPOSITION XXV. THEOREM. 108. Tuo triangles are equal when the three sides of the one are equal respectirely to the three sides of the other.

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Hyp

In the triangles A B C and A' B' C', let A B = A' B',

AC = A' C', BC = B'C'.
We are to prove A ABC = A A'B'C'.

Place A A' B' C' in the position A B' C, having its greatest side A' C' in coincidence with its equal A C, and its vertex at B', opposite B.

Draw B B' intersecting A C at H.
Since A B = A B',

Hyp. point A is at equal distances from B and B'.

Since BC= B' C,
point C is at equal distances from B and B'.

.. A C is I to B B' at its middle point, $ 60 (two points at equal distances from the extremities of a straight line deter

mine the I at the middle of that line). Now if A A B' C be folded over on AC as an axis until it comes into the plane of A ABC,

H B' will fall on H B,
(for 2 A H B = L A H B', each being a rt. Z),
and point B' will fall on B,

(for H B' = H B).
.. the two A coincide, and are equal in all respects.

Q. E. D.

PROPOSITION XXVI. THEOREM. 109. Two right triangles are equal when a side and the hypotenuse of the one are equal respectively to a side and the hypotenuse of the other.

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In the right triangles A B C and A' B'C', let A B=A' B',

and AC = A'C'.

We are to prove A A BC= A A' B'C'.

Take up the A A B C and place it upon A A' B'C', so that A B will coincide with A' B'.

Then BC will fall upon B'C',
(for Z A BC= 2 A' B' C', each being a rt. 2),

and point C will fall upon C"; otherwise the equal oblique lines A C and A' C' would cut off unequal distances from the foot of the I, which is impossible,

§ 57 (two equal oblique lines from a point in a I cut off equal distances from the

foot of the I).
.. the two o coincide, and are equal in all respects.

Q. E. D.

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110. Two right triangles are equal when the hypotenuse and an acute angle of the one are equal respectively to the hypotenuse and an acute angle of the other.

AI

- BIL

In the right triangles A B C and A' B'C', let A C= A' C',

and ZA = A'.
We are to prove A ABC= A A' B'C'.
A C= A' C',

Hyp.
ZA= ZA',

Hyp. then <C=LC',

§ 101 (if two rt. A have an acute 2 of the one equal to an acute < of the other,

then the other acute & are equal).
i. A A BC= A A' B'C',

§ 107 (two S are equal when a side and two adj. £ of the one are equal respectively to a side and two adj. of the other).

Q. E. D.

111. COROLLARY. Two right triangles are equal when a side and an acute angle of the one are equal respectively to an homologous side and acute angle of the other.

PROPOSITION XXVIII. THEOREM.

112. In an isosceles triangle the angles opposite the equal sides are equal.

Let A B C be an isosceles triangle, having the sides

AC and C B equal.

We are to prove ZA= 2 B.

From C draw the straight line CE so as to bisect the ZA CB.

In the A AC E and BCE,
A C=BC,

Hyp.
CE=CE,

Iden.

ZACE= ZBCE;

Cons. .. A ACE=A BCE,

§ 106 (two A are equal when two sides and the included of the one are equal respectively to two sides and the included 2 of the other).

..ZA= LB,
(being homologous É of equal A).

Q. E. D.

Ex. If the equal sides of an isosceles triangle be produced, show that the angles formed with the base by the sides produced are equal.

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