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185. Two parallelograms, having two sides included angle of the one equal respectively sides and the included angle of the other, ar

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In the parallelograms ABCD and A'B'C'D', A'B', AD= A'D', and ▲ A = ▲ A'.

To prove that the [$] are equal.

Apply □ ABCD to □ A'B'C'D', so that AD wi and coincide with A'D'.

Then AB will fall on A'B',

(for A=A', by hyp.),

and the point B will fall on B',
(for AB A'B', by hyp.).

=

Now, BC and

B'C' are both

to A'D' and ar

through point B'.

.. the lines BC and B'C' coincide,

and C falls on B'C' or B'C' produced.

In like manner, DC and D'C' are to A'B' and ar through the point D'.

.. DC and D'C' coincide.

.. the point C falls on D'C', or D'C' produced .. C falls on both B'C' and D'C'.

.. C' must fall on the point common to both, namel .. the two coincide, and are equal.

186. COR. Two rectangles having equal bases and are equal.

187. If three or more parallels intercept equal parts on any transversal, they intercept equal parts on every transversal.

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Let the parallels AH, BK, CM, DP intercept equal parts HK, KM, MP on the transversal HP.

To prove that they intercept equal parts AB, BC, CD on the transversal AD.

Proof. From A, B, and C suppose AE, BF, and CG drawn. Il to HP.

Then AE= HK, BF= KM, CG
(parallels comprehended between parallels are equal).

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(being ext.-int. ▲ of || lines);

LE LF LG,

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MP,

§ 180

Ax. 1

§ 106

§ 112

$ 147

(having their sides || and directed the same way from the vertices).

ΔΑΒΕ Δ BCF = Δ CDG,

(each having a side and two adj. 4 respectively equal to a side and two

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D

bisecting one side, bisects the other side also. For, let DE be I to BC and bisect AB. Draw through A a line to BC. Then this line is to DE, by § 111. The three parallels by hypothesis intercept equal parts on the transversal AB, and there- B fore, by $187, they intercept equal parts on the t AC; that is, the line DE bisects AC.

189. COR. 2. The line which joins the middle po sides of a triangle is parallel to the third side, and is half the third side. For, a line drawn through D, th point of AB, I to BC, passes through E, the middle AC, by § 188. Therefore, the line joining D and E with this parallel and is to BC. Also, since ER to AB bisects AC, it bisects BC, by § 188; that is, =BC. But BDEF is a □ by construction, and DE=BF=BC.

D

190. COR. 3. The line which is parallel to the bases ezoid and bisects one leg of the trapezoid bisects the other leg also. For if parallels intercept equal parts on any transversal, they intercept equal parts on every transversal by § 187.

A

E

F

191. COR. 4. The median of a trapezoid is parallel to the bases, and is equal to hal of the bases. For, draw the diagonal DB. In the join E, the middle point of AD, to F, Then, by § 189, EF is I to AB and join Fto G, the middle point of BC.

=

and DC. AB and FG, being II to

=

the middle poir

AB. In the

Then FG is

DC, are Il to ea

But only one line can be drawn through Fl to AB. fore FG is the prolongation of EF. Hence EFG is and DC, and } (AB+DC).

=

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30. The bisectors of the angles of a triangle meet in a point which is equidistant from the sides of the triangle.

HINT. Let the bisectors AD and BE intersect at O.
Then O being in AD is equidistant from AC and AB.
(Why?) And O being in BE is equidistant from BC
and AB. Hence O is equidistant from AC and BC, A
and therefore is in the bisector CF.

(Why?)

B

31. The perpendicular bisectors of the sides of a triangle meet in a point which is equidistant from the vertices of the triangle.

HINT. Let the bisectors EE' and DD' intersect at O. Then O being in EE' is equidistant from A and C. (Why?) And O being in DD' is equidistant

from A and B.
is in the

E

D

E

D'

B

F

Hence O is equidistant from B and C, and therefore bisector FF'. (Why?)

32. The perpendiculars from the vertices of a triangle to the opposite sides meet in a point.

(why ?), and AB' = BC, and AC'

B

A

P

K

H

HINT. Let the s be AH, BP, and CK. Through A, B, C suppose B'C', A'C', A'B' drawn to BC, AC, AB, respectively. Then AH is to B'C'. (Why?) Now ABCB' and ACBC are = BC. (Why?) That is, A is the middle point of B'C'. In the same way, B and C are the middle points of A'C' and A'B', respectively. Therefore, AH, BP, and CK are the bisectors of the sides of the ▲ A'B'C'. Hence they meet in a point. (Why?)

H

D

33. The medians of a triangle meet in a point which is two-thirds of the distance from each vertex to the middle of the opposite side. HINT. Let the two medians AD and CE meet in O. Take Fthe middle point of OA, and G of OC. Join GF, FE, ED, and DG. In ▲ AOC, GF is to AC and equal to § AC. (Why?) DE is | to AC and equal to AC. (Why?) Hence DGFE is a □. Hence AF FOOD, and CG = GO = OE. Hence, any median cuts off on any other median two-thirds of the distance from the vertex to the middle of the opposite side. Therefore the

(Why?)
(Why?)

A

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192. A polygon is a plane figure bounded by stra The bounding lines are the sides of the polygon, sum is the perimeter of the polygon.

The angles which the adjacent sides make with are the angles of the polygon, and their vertices ar tices of the polygon.

The number of sides of a polygon is evidently eq number of its angles.

193. A diagonal of a polygon is a line joining th of two angles not adjacent; as AC, Fig. 1.

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194. An equilateral polygon is a polygon which sides equal.

195. An equiangular polygon is a polygon which angles equal.

196. A convex polygon is a polygon of which no si produced, will enter the surface bounded by the peri

197. Each angle of such a polygon is called a sali and is less than a straight angle.

198. A concave polygon is a polygon of which tw sides, when produced, will enter the surface bounde perimeter. Fig. 3.

199. The angle FDE is called a re-entrant angl greater than a straight angle.

If the term polvgon is used, a convex polygon is m

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