Ex. 674. Construct a triangle that shall have a given perimeter and shall be similar to a given triangle.

Ex. 675. Construct a trapezoid, given the two bases and the two diagonals. HINT. How do the diagonals of a trapezoid divide each other?

PROPOSITION XVI. THEOREM

428. If two triangles have an angle of one equal to an angle of the other, and the including sides proportional, the triangles are similar.

1. Place A DEF on ▲ ABC so that D shall

2.

coincide with ▲ A, DE falling on AB,
and DF on AC. Represent ▲ DEF
in its new position by A AHK.

AB AC

REASONS

1. § 54, 14.

2. By hyp.

6. But A AHK is ▲ DEF transferred to a

F

Ex. 676. Two triangles are similar if two sides and the median drawn to one of these sides in, one triangle are proportional to two sides and the corresponding median in the other triangle.

=

Ex. 677. In triangles ABC and DBC, Fig. 1, AB AC and BD = BC. Prove triangle ABC similar to triangle DBC.

=

Ex. 678. In Fig. 2, AB: AC AE: AD. Prove triangle ABC similar to triangle ADE.

Ex. 679. If in triangle ABC, Fig. 3, CA = BC, and if D is a point such that CA: AB = AB: AD, prove AB

Ex. 680.

BD.

Construct a triangle similar to a given triangle and having

the sum of two sides equal to a given line.

PROPOSITION XVII. THEOREM

429. Two triangles that have their sides parallel each to each, or perpendicular each to each, are similar.

Given

AABC and A'B'C', with AB, BC, and CA || (Fig. 1)

or (Fig. 2) respectively to A'B', B'C', and C'A'.

1. AB, BC, and CA are or respectively to A'B', B'C', and c'A'.

2... A, B, and C are equal respectively or

are sup. respectively to A', B', and c'.

REASONS 1. By hyp.

2. §§ 198, 201.

ARGUMENT

3. Three suppositions may be made, there-
fore, as follows:

(1) ▲ ▲ + 2 A' = 2 rt. ≤, ≤ B + ≤ B'
=2rt. 4, C+ <c = 2 rt. .
(2) ZA=ZA', ZB + Z B' = 2 rt. 4,
<c+c=2 rt. .

s.

4. According to (1) and (2) the sum of the

s of the two A is more than four rt. s.

5. But this is impossible.

6. .. (3) is the only supposition admissible; i.e. the two A are mutually equiangular.

REASONS

3. § 161, a.

7. .. AABC ~ AA'B'C'.

~

Q.E.D.

7. § 420.

430. Question. Can one pair of angles in Prop. XVII be supplementary and the other two pairs equal?

SUMMARY OF CONDITIONS FOR SIMILARITY OF TRIANGLES

431. I. Two triangles are similar if they are mutually equiangular.

(a) Two triangles are similar if two angles of one are equal respectively to two angles of the other.

(b) Two right triangles are similar if an acute angle of one is equal to an acute angle of the other.

(c) If a line is drawn parallel to any side of a triangle, this line, with the other two sides, forms a triangle which is similar to the given triangle.

II. Two triangles are similar if their sides are proportional. III. Two triangles are similar if they have an angle of one equal to an angle of the other, and the including sides proportional.

IV. Two triangles are similar if their sides are parallel each to each, or perpendicular each to each.

Ex. 681. Inscribe a triangle in a circle and circumscribe about the circle a triangle similar to the inscribed triangle.

Ex. 682. Circumscribe a triangle about a circle and inscribe in the circle a similar triangle.

Ex. 683. The lines joining the mid-points of the sides of a triangle form a second triangle similar to the given triangle.

Ex. 684. ABC is a triangle inscribed in a circle. A line is drawn from A to P, any point of BC, and a chord is drawn from B to a point Q in arc BC so that angle ABQ equals angle APC. Prove AB × AC = AQ × AP.

PROPOSITION XVIII. THEOREM

432. The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the other two sides.

Given AABC with BP the bisector of ▲ ABC.

To prove AP: PC = AB: BC.

ARGUMENT

1. Through a draw CE || PB, meeting AB

5. By hyp.
6. § 54, 1.

7. § 162.

8.

.. AP PC AB: BC.

Q.E.D.

8. § 309.

Ex. 685. The sides of a triangle are 8, 12, and 15. ments of side 8 made by the bisector of the opposite angle.

Find the seg

Ex. 686. In the triangle of Ex. 685, find the segments of sides 12 and 15 made by the bisectors of the angles opposite.

PROPOSITION XIX. THEOREM

433. The bisector of an exterior angle of a triangle divides the opposite side externally into segments which are proportional to the other two sides.

Given ▲ ABC, with BP the bisector of exterior CBF.

To prove AP PC AB: BC.

Ex. 687. Compare the lettering of the figures for Props. XVIII and XIX, and also the steps in the argument. Could one argument serve for the two cases?

Ex. 688.

The sides of a triangle are 9, 12, and 16. Find the segments of side 9 made by the bisector of the exterior angle at the opposite vertex.