Proposition 23. Theorem.

108. Two triangles are equal if the three sides of the fone are equal respectively to the three sides of the other. Hyp. Let ABC, DEF be

=

two As having AB DE, AC DF, BC = EF.

To prove ▲ ABC = A DEF. Proof. Apply the ▲ ABC to the A DEF so that the side AB shall coincide with its equal DE, and the vertex C fall at F' on the opposite side of DE from F.

Join FF' which cuts DE at H.

A

F

Because

DFDF', and EF EF',

=

(Hyp.)

.. points D and E are equally distant from F and F'.

.. DE is to FF' at its middle pt. H.

(67)

Now revolve the A DEF about DE as an axis till it comes into the plane of the ▲ DEF'.

... the two As coincide in all their parts, and are equal.

Q. E.D.

109. SCH. When two triangles are equal, the equal angles lie opposite the equal sides; and conversely, the equal sides lie opposite the equal angles. Thus, the angles A and D are equal or homologous angles (95). Also the angles C and F are homologous; likewise the angles B and E.

Proposition 24. Theorem.

110. Two right-angled triangles are equal when the hypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other.

Hyp. Let ABC, DEF be two rt. As having

E

Proof. Apply the A ABC to the A DEF so that BC shall coincide with its equal EF.

Then,

since B ZE = art. 2,

.. BA will lie along ED, and pt. A will fall somewhere on ED.

But the equal oblique lines CA and FD cut off on ED equal distances from the foot of the

EF (61).

.. pt. A will fall at D.

.. the two As coincide in all their parts, and are equal.

EXERCISES.

Q.E.D.

1. If BC, the base of an isosceles triangle, ABC, is produced to any point D, show that AD is greater than either of the equal sides.

2. Prove that the sum of the distances of any point from the three vertices of a triangle is greater than half its perimeter.

3. If one of the acute angles of a right triangle is 40° 14′ 48′′, what is the value of the other acute angle?

Proposition 25. Theorem.

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

Hyp. Let ABC be an isosceles ▲ having

Proof. Draw the line CD from the

vertex C, to the middle pt. D of the A

CD CD, (being common to both As,)

.. A ADC = A BDC.

Two As are equal if the three sides are equal each to each (108).

.. ZA = < B,

being opposite equal sides (109).

Q.E.D.

112. COR. 1. Since ▲ ADC = ▲ BDC, we have ACD = BCD, and ADC = BDC. Therefore, the straight line which joins the vertex to the middle point of the base of an isosceles triangle, is at right angles to the base, and bisects the vertical angle.

Hence, also, the bisector of the vertical angle of an isosceles triangle bisects the base at right angles.

113. COR. 2. The perpendicular from the vertex to the base of an isosceles triangle bisects the base and the angle at the vertex.

An equilateral triangle is also equiangular.

Proposition 26. Theorem.

114. Conversely, if two angles of a triangle are equal, the sides opposite the equal angles are equal, and the triangle is isoceles.

Hyp. Let ABC be a ▲ having

having a side and the two adjacent angles equal, each to each (105).

... AC

=

BC,

being homologous sides of equal ▲s (109).

Q.E.D.

115. COR. An equiangular triangle is also equilateral.

EXERCISES.

1. ABC is an isosceles triangle, with AB = AC. The bisectors of the angles B and C meet at O.

Prove that

2. ABC is a triangle; BA is produced to D so that AD AC, and DC is joined. Prove that BCD > < BDC. 3. The angle C is twice as large as either of the angles A and B: how many degrees are there in each angle?

Proposition 27. Theorem.

116. If one side of a triangle be greater than another, the angle opposite the greater side is greater than the angle opposite the less.

A

Hyp. Let ABC be a ▲ having the side

But ext. CDB > opposite int. DAB of ▲ ADB. (98)

1. Prove Prop. 27 by producing CB to E, making CE = CA.

2. ABC is a triangle in which OB, OC bisect the angles B, C, respectively; show that, if AB is greater than AC, then OB is greater than OC,