PROPOSITION XVI. THEOREM.

292. Two triangles are similar, if they have their sides respectively parallel or perpendicular to each other.

Given Triangles ABC, A'B'C', having their sides respectively parallel or respectively perpendicular to each other;

To Prove Triangle ABC is similar to triangle A'B'C'.

Since their sides are || or to each other, (Hyp.)

any two homologous angles of these triangles must be either equal or supplementary.

(116, 118)

Hence there are three conceivable cases to consider.

1o. Suppose all the angles of the one triangle respectively supplemental to the homologous angles of the other.

i.e., A+ A'= a st. ≤, B+ B'= a st. ≤, and also C + c' a st. Z.

2o. Suppose two supplemental and one equal.

i.e., A = A', B + B'= a st. ≤, and c + c'= a st. Z.

3°. Suppose all the angles are mutually equal.*

i.e., A = A', BB', and ... C = c'.

Each of the first two suppositions must be rejected, since

*If two are right angles, they are equal and supplemental.

the sum of the angles of two triangles cannot exceed two straight angles. Hence the third alone is admissible;

.. AABC is similar to ▲ A'B'C'. Q.E.D. (287) 293. SCHOLIUM. The homologous sides are those mutually parallel or perpendicular.

PROPOSITION XVII. THEOREM.

294. Two polygons are similar if composed of the same number of triangles similar each to each and similarly placed.

E

A

B

Β'

Given: In polygons P and P', triangles AED, ADC, ACB, similar to triangles A'E'D', A'D'C', A'C'B', respectively, and similarly placed;

1o. P and P' are mutually equiangular.

For their homologous are either homologous of similar ▲, or are like sums of homologous

lar A;

of simi(Hyp.)

.. LE = LE', ≤ EDA + ADC = E'D'A' + ZA'D'C', etc.

2o. The homologous sides of P and P' are proportional.

For since the A are similar,

(Hyp.)

Q.E.D. (284)

AB: A'B' = BC: B'C' = AC : A'c' = CD : C'D', etc.;

.. P is similar to P'.

PROPOSITION XVIII. THEOREM.

295. Conversely, two similar polygons may be divided into the same number of triangles, similar to each other and similarly placed.

E

D

B

Given: ABCDE, or P, and A'B'C'D'E', or P', two similar polygons;

To Prove: P and P' may be divided into the same number of similar triangles similarly placed.

From 4 and 4', homologous vertices of P and P', draw diagonals AC, AD, and A'C', A'D', respectively.

1o. The number of triangles thus formed in P and P', respectively, is the same. For it is equal to the number of sides in each, less two.

2o. These triangles are similar and similarly placed.

Again, since A AED is similar to A A'E'D', (Above)

ED: DA E'D': D'A'.

=

But ED: DC = E'D' : D'C',

(285)

(Hyp.)

(249)

(290)

.. DA: DC = D'A' : D'C',

.. AADC is similar to A A'D'C'.

In the same way, the remaining triangles of P may be proved similar to the similarly placed triangles in P'.

.. P and P' may be divided as stated.

Q.E.D.

PROPOSITION XIX. THEOREM.

296. The perimeters of two similar polygons have the same ratio as any two homologous sides.

D

E

B

Given: AB, A'B', any two homologous sides of two similar polygons, P, P', of which p and p' are the perimeters;

To Prove:

p: p' = AB: A'B'.

Since P and P' are similar polygons,
AB: A'B' = BC: B'C' = CD = C'D', etc.,

.. ABBC + CD + ••• : A'B' + B'C' + C'D' +

(Hyp.) (284)

··· = AB : A'B'.

(251)

But p=AB+BC+CD+..., and p' A'B'+B'C'+C'D'+···,

=

.. p: p' = AB : A'B'.

Q.E.D.

EXERCISE 361. Draw a diagram to show that two figures may be mutually equiangular though not similar; and another to show that two figures may have their sides mutually proportional and yet not be similar.

362. In the diagram for Prop. XVII., if the numerical measures of BC and B'C' are, respectively, 16 and 10, what is the ratio of p to p'?

RATIOS OF CERTAIN LINES.

PROPOSITION XX. THEOREM.

297. In a right triangle, if a perpendicular be drawn from the vertex of the right angle to the hypotenuse,

1°. The perpendicular is a mean proportional between the segments of the hypotenuse.

2o. Each arm is a mean proportional between the hypotenuse and the adjacent segment.

Given: In a right triangle ABC, AD perpendicular to BC from the right angle BAC;

To Prove:

{

2°. BC: AB=AB:BD, and BC: AC=AC: DC.

Since BDA and BAC are rt. A,
and acute B is common to both,
rt. ▲ BDA is similar to rt. ▲ BAC.
Since CDA and BAC are rt. A,
and acute ≤ C is common to both,
rt. A CDA is similar to rt. ▲ BAC,
.. rt. ▲ BDA is similar to rt. ▲ CDA,
each being similar to rt. ▲ BAC.
Since ▲ BDA is similar to ▲ CDA,

BD: DADA: DC.

Q.E.D.

Since ABAC is similar to A BDA and CDA,

(Hyp.)

(288) (Hyp.)

(288)

(286)

(284)