PROPOSITION XI.

§ 85. Given, between A and BC, AD BC, AE and AF meeting BC at equal distances from D, and AC meeting BC at a greater distance from D than AE or AF.

B E

Required, AD AE≈≈ AF ≈ AC.

XXXVI.

(1.) In the right triangle ADE, ɑAED ≈≈ ɑADE?

AD AE?

73, 75. e.

(2.) Can you show that the triangles ADE and ADF are

identical?

AE AF?

39. I.

(3.) In tAFC, aACF ≈≈≈ aAFC? Why?

AF (or its equal AE) ≈≈≈ AC?

§ 86. THEOR. XI. Of the lines that can be drawn to a given straight line from a point without, (1.) The perpendicular is the shortest.

(2.) Of oblique lines, those which meet the given line at equal distances from the perpendicular are equal; and (3.) those which are nearer the perpendicular are shorter than those which are more remote.

[Proved by applying Theor. IX., &c.]

§ 87. a.) Hence a perpendicular is the measure of distance between a point and a straight line. Compare § 83.

88. b.) The converse of this Theorem may be readily shown; thus,

(1.) If AD is the shortest line that can be drawn from A

* I

to BC, it is BC; for, otherwise, a perpendicular could be drawn which would be shorter.

(2.) If AE

=

=

AF, then DE DF; for, if DE were >

or < DF, AE must be > or < AF.

(3.) If AE <AC, then DE < DC; for if DE were = or > DC, AE must be : or > AC.

§ 89. c.) Having drawn, from A to BC, the perpendicular AD, draw any other line AE. Show, by comparing the angles ADE and AED, that AE cannot be BC. See $ 50.

Show the same, by comparing the exterior angle AEB with the interior opposite angle ADE.

How many lines can be drawn from A to BC, making the two adjacent angles equal?

As an oblique line between A and BC is drawn farther from the perpendicular, does it make the adjacent angles more or less unequal (or, in other words, does it meet BC more or less obliquely)?

§ 90. d.) Besides AE, can you draw another line from A to BC, equal to AF? How many lines can you draw from A to BC, equal to AC?

Can two lines drawn from A to BC upon the same side of AD be equal to each other?

Can, then, three equal lines be drawn from A to BC?

COR. I. Of the straight lines that can be drawn to a given straight line from a point without, —(1.) no three can be equal; and (2.) any two that are equal must be upon opposite sides of the perpendicular.

§ 91. e.) Given, any point D in DE drawn AB through C, the middle point of AB; and any point F, not in DE.

Required, DA DB; and FA ~FB.

As DA and DB meet AB at equal distances from the perpendicular DC, ᎠᎪ :: ᎠᏴ ?

A

XXXVII.

D

GF

E

86.2.

From G (where FA crosses the perpendicular), draw GB.

COR. II. If through the middle point of a given straight line a perpendicular be drawn, any point in that perpendicular will be equally distant from the two extremities of the given line; and any point without the perpendicular will be unequally distant.

§ 92. f.) Given, in DE (Fig. XXXVII.), any two points D and G, each equally distant from the two extremities of AB.

Required, DE AB; and AC AB.

If no point not in the perpendicular drawn through the middle of AB can be equally distant from A and B, then in what line must the points D and G both lie?

91.

Can any other straight line pass through them both, except DE?

DE AB? and AC CB?

6. a.

COR. III. If any two points of one straight line are each equally distant from the extremities of a second, the first line will bisect the second at right angles.

g.) Hence, if one straight line passes through the middle of a second, and any other point of the first is equally distant from the extremities of the second, what angles do the two lines make with each other?

§ 93. h.) In an isosceles triangle,

(1.) How does a perpendicular drawn from the vertex to the base divide the base?

88.2.

(2.) What angles does a line

XXVI.

A

drawn from the vertex to the mid- B

dle of the base make with the base?

92. g.

(3.) If a perpendicular be raised from the middle of the base, will it pass through the vertex? Why?

91

In any triangle not isosceles, will a perpendicular raised from the middle of the base pass through the vertex?

If ADB and AGB (Fig. XXXVII.) are two isosceles triangles upon the same base AB, how will the straight line DE passing through their vertices divide the base? at what angles?

Of the two sides AB and AC, let AB be that which is not longer than the other. Then place tABD upon tABC

As AD = AC, must pD fall without tABC?

Join CD. Then, as AC AD, aADC ≈ aACD?

But

=

BDC ADC, and ACD ≈≈≈ BCD?

BDCBCD?

BCBD?

73. a.

20. C.

73.

§ 95. II. Given, in tABC and tABD (Fig. XXXVIII.), AB = AB, and AC = AD, but

96. THEOR. XII. In triangles agreeing in two sides, the greater included angle is subtended by the greater third side.

[Proved by superposition with the aid of Theor. IX., and by reductio ad absurdum.]

97. REMARK. In § 94, we might proceed thus, without joining CD:

AFCF AC, and FD + FB ≈≈ BD?

.. AD(=AF+FD)+BC(=CF+FB)

78.

AC+BD? 20. e.

20.g.

Can you now prove that the two triangles are identical? 39. 1.

Name each pair of equal angles.

A.) Triangles agreeing in the three sides are identical.

§ 99. II. Given, two triangles ABC and DEF (Fig. IX., § 98) agreeing in two angles and the side opposite to one of them; viz. aA = aD, aB aE, and BC= EF.

=

Required, tABCDEF.