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Theorem 5-1Opposite sides of a parallelogram are congruent

Reason: since the sides are parallel and they meet in equidistant

Theorem 5-2

Opposite angles of a parallelogram are congruent.

Reason: since it has 2 pairs of parallel congruent sides which

divides 360 degrees in also 2 pairs of congruent angles.

Theorem 5-3

Diagonals of a parallelogram bisect each other.

Reason: since they have equal sides and equal angles

Theorem 5-4

If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

Reason: since a parallelogram is a quadrilateral havingboth pairs of opposite sides

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Theorem 6-6

(SSS Inequality Theorem)

If two sides of one triangle are congruent to two sides of another triangle, but the third side of the firsttriangle is longer than the third side of the second, then the included angle of the first triangle is larger

than the included angle of the second.

By the SSS inequality theorem, segment BC is longer than segment EF.

Reason: the included angle of the first triangle is larger than the second triangle

because it has longer hypotenuse.

Theorem 7-1

If the measures of two sides of a triangle are proportional to the measures of two corresponding sides ofa second triangle and the included angles are congruent then the triangles are similar

IfAB = f EF

and

BC= f FG

andB =F

thenABC =EFG

Reason: the two triangles have congruent 2 sides and included angle, so they have congruenthypotenuse. Therefore, they are similar.

Theorem 8-1

If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar tothe original triangle and to each other

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Triangle ABC, triangle ADB, and triangle BDC are all similar

Corollary I

When the altitude is drawn to the hypotenuse of a right triangle, the length of the altitude is the geometricmean between the segments of the hypotenuse.

AD/BD=BD/CD

Corollary II

When the altitude is drawn to the hypotenuse of a right triangle, each leg is the geometric mean between

the hypotenuse and the segment of the hypotenuse that is adjacent to that leg.

AC/AB=AB/AD

AC/CB=CB/BC

Theorem 9-11

If a secant and a tangent intersect at the point of tangency, then the measure of each angel formed isone-half the measure of its intercepted arc.

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Theorem 9-13

If two secants, a secant and a tangent, or two tangents intersect in the exterior of a circle, then themeasure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.

Example:

Use k to find the value ofx.

Two secants that intersect outside of the circle:

This is one of 3 cases that we get from Theorem 9-13. It tells us that the measure of the angle formed by

these secants is half of the distance between the measures of the 2 intercepted arcs. This difference will

ALWAYS be positive.

Theorem 11-7

If the scale factor of two similar figures is a:b, then

(1) the ratio of the perimeters is a:b.

(2) the ratio of the areas is a2:b2

Example:

Find the ratio of the perimeters and the ratio of the areas of the ratio of the areas of the twosimilar figures.

Solution:

The scale factor is 8:12, or 2:3. Therefore, the ratio of the perimeters is 2:3. The ratio of theperimeters is 2:3. The ratio of the areas is 22 : 32 or 4 : 9.

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Theorem 12-7

The Lateral area of a cone equals half the circumference of the base times the slant height.

(L.A.= 1/2(2 r)l) or (L.A.= rl)

The height: 8

The slant height:10

The radius: 5

Lateral Area= r l = (5)(10)=500

Theorem 12-8

The Volume of a cone equals one-third the area of the base times the height of the cone. (V= 1/3 r2h)

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The height: 8

The slant height:10

The radius: 5

Lateral Area= r l = (5)(10)=500

Volume= 1/3 r2h =1/3 (52)(8)= 200/3

Theorem 13-1

The Distance Formula

The distance d between points (x1, y1) and (x2, y2) is given by:

Find the distance between the points.

1. (3, 8), (2, 4) =412. (1, 8), (5, 3) =157

Theorem 13-2

An equation of the circle with center (a, b) and radius r is:

(x a)2 + (y b)2 = r2

Example:Find the center and the radius of the circle with equation

(x 1)2 + (y + 2)2 =9. Sketch the graph.

Solution:(x 1)2 + (y + 2)2 =32

The center point is (1, 2) and the radius is 3.

2

12

2

12 )()( yyxxd +=

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The graph is shown below.

Theorem 14-8

A composite of reflections in two intersecting lines is a rotation about the point of intersection of the twolines. The measure of the angle of rotation is twice the measure of the angle from the first line of

reflection to the second.

Given:

j intersects k, forming an angle of measure y at O.

Proof:

The diagram shows an arbitrary point P and its image P by reflection in j. According to the definition of a

rotation we must prove that OP = OP and mPOP= 2y.

Rj and Rk are isometries, so they preserve both distance and angle measure. Therefore OP= OP, OP =

OP, m1= m2, and m3 = m4. Thus OP = OP and the measure of the angle of rotation equals:

m1 + m2 + m3 + m4=2m2 + 2m3= 2y.

Corollary

A composite of reflections in perpendicular lines is a half-turn about the point where the lines intersects.

References:library.thinkquest.com and google.com and yahoo.com