Download MATH FINAL (editted).

MATH FINAL
Please note that throughout this file << means angle
O
B
Circles and angles inside circles:
A
 Arc: smallest arc between two points. An arc is measured in radians.
A
ɣ
 <<AB (arc) = αr
 α length = arc<<AB
r
 360o = 2; 1o = /180o
ɣ
2ɣ
α
B
O
r
 Circumference (C) of circle = 2R
 Central Angle and Inscribed Angle:
B
O 2δ
r
β
δ
r
 Being that O is the center of the circumference:
δ
δ
 α = central angle; α length = arc<<AC; α = 2β
C
δ
- the proof: if a line is drawn from point B to point O (forming a radius) then β is divided into
ɣ and δ. The two triangles formed are both isosceles since two of their sides are radii (which
are always congruent.) The external angle O equals either 2ɣ or 2δ the sum of
2ɣ + 2δ = α thereby we know that α = 2β
 β = inscribed angle; β = α/2; β = (arc<<AC)/2
 If there are a few inscribed angles all of whose endpoints lie on the two endpoint of
the same intercepted arc than they are congruent  <<A and <<B in this picture are
P
congruent to each other
β
 In every quadrilateral that can be inscribed the opposite angles are supplementary (the
sum of the two angles = 180o)
360 -2β
O
 Interior Eccentric Angle:
Exterior Eccentric Angle: A
2β
B
A
A
<<AB + <<CD
<<AB - <<CD
180 -β
P
C
α = ----------------β = ----------------C
α
2
2
Q
B
D
B
β
 Circumscribed Angle:
Segment Angle:
D
<<ACB - <<AB
α = ----------------------2
<<AB
β = -----2
A
C
180-2(90 -β) = 2β
O
O
α
β
B
2β
r
90 -β
r
2β
B
P
A
 Thales Theorem: if A,B, and C are points on a circle where the line AC is a diameter 90 -β
of the circle, then the angle ABC is a right angle
α
 Proof: AC=CB=CD=radius, thereby triangle OBC is isosceles, and triangle AOB is
also isosceles. Triangle ABC = α +(α+ β) + β = 180o  2α + 2β = 180o  α +β = 90o
αβ
β
Basic Trigonometry
 The following is a simple chart with the trig that can help us on the final
 In order to remember how to use sine, cosine, and tangent simply remember
the acronym SOHCAHTOA – which is deciphered: Sine is Opposite over
Hypotenuse; Cosine is Adjacent over Hypotenuse; Tangent is Opposite over
Adjacent.
Congruence & Similarity of Triangles
 Triangles can be congruent by: SSS (side-side-side), ASA (angle-side-angle), SAS (side-angle-side), SAA
(side-angle-angle), and HL (height-leg – if there is a right angle where the height of the triangles and the
leg attached to the 90o angle are congruent then the two triangles are congruent)
δ

Note: triangles can not be congruent by AAA (angle-angle-angle) or by SSA (side-side-angle – unless
the angle is a right angle)
 Triangles can be similar (all angles congruent and side a proportional to each other) by: AA (angle-angle –
if two angles are congruent the third will also be congruent), SAS (side-angle-side), SSS (side-side-side),
HL (height-leg – if there is a right angle where the height of the triangles and the leg attached to the 90o
angle, where the corresponding sides have identical ratios),
 Note: SSA (side-side-angle – unless the angle is a right angle)
The relationship between a point P inside a circumference
 P is internal:
P is external:
A
(PA)(PB) = (PC)(PD)
(PA)(PB)=(PC)(PD)
C
P
D
B
Proof:
<<APD is congruent to <<CPB (opposite by same vertex)
<<DAB is congruent to <<BCD (inscribed angle by arc DB)
Triangle APD is thereby similar to triangle CPB by AA
So the ratio would be:
PA = PC
PD PB
This means that (PA)(PB) = (PC)(PD)
B
A
P
C
D
Proof:
<<BPD is congruent to <<DPB (same angle)
<<ADC is congruent to <<CBA (inscribed angle by arc AC)
Triangle BCP is thereby similar to triangle DAP by AA
So the ratio would be:
PA = PC
PD PB
This means that (PA)(PB) = (PC)(PD)
(PM)2 = (PN)2 PM = PN
(PA)(PB) = (PM)2
M
M
P
O
A
P
B
N
Proof:
<<MPB is congruent to <<MPA (same angle)
<<PMA is congruent to <<MBA (PMA is a segment
angle and MBA is an inscribed angle both of the same
arc MA)
Triangle PMB is thereby similar to triangle PAM by AA
So the ratio would be:
PM = PB
PA PM
This means that (PA)(PB) =A(PC)(PD)
D
A+C=B=D
C
Proof:
<<OMP =90o definition of tangent line
<<ONP =90o definition of tangent line
<<OMP = <<ONP  if two angles are rt. << then they’re congruent
OM = ON  radii of same circle are congruent
OP =OP  reflexive property
Triangle POM is congruent to triangle PON  SSA
PM = PN  corresponding parts of congruent triangles are congruent
B
Cevians of Triangles: line segments where a vertex forms one of the endpoints and the opposite side is
another endpoint
 Median – the line segment going from one vertex to the midpoint on the opposite side
 Centroid or center of gravity – the C.G. of a triangle is the point of intersection of the medians of this
triangle [property: the C.G. divides the median where from C.G. to vertex is twice the length as from
C.G. to opposite side – thus forming a 2:1 ratio.]
 Altitude – the line segment going from one vertex to the opposite side making a 90o angle [an acute triangle
has 3 altitudes, a right triangles has 1 altitude that is not part of the triangle (the other two altitudes would
be the two legs of the triangle) an obtuse triangle as 1 altitude.]
 Orthocenter – the orthocenter of a triangle is a point of intersection of all altitudes [the orthocenter is
only interior to the triangle when its acute]
 Internal bisectors – segments that split in half the angle of the vertex and has the opposite side as its
endpoint [every triangle has 3 internal bisectors]
 Incenter – the incenter of a triangle is the point of intersection of all internal bisectors [the incenter is
always interior to the triangle, and it is also the center of a circumference inscribed in a triangle]
 Remarks about the points of intersection:
 If triangle ABC is isosceles, then the three points of intersection are collinear
 If triangle ABC is equilateral then the three points coincide
Areas
Parallelogram: A = bh
Square: A=l2
b
l
h
l
Rectangle: A = (width)(height)
Trapezoid:
Triangle: A = bh/2
Rhombus: A = ½(d1d2)
h
b
b
b
radius
 The radius of a regular polygon is the radius of a circumscribed
circumference – it bisects the angle of the polygon
apothem
 The apothem of a regular polygon is the radius of the inscribed
circumference
 All apothems of regular polygons are congruent
 Only regular polygons have apothems
 An apothem forms a right angle with a side
 The area of a regular polygon is A= ½(apothem)(perimeter)
 The area of a regular hexagon is A = ½(s2(route3))
 To compare ratio of areas:
 Since the ration of the sides is x/y and the height changes the ratio of the smaller
x
triangle to the bigger triangle = x2/y2  the area of the trapezoid formed would
be x2/(y2 - x2)
y
In a triangle the mid-line = ½ the length of the base
In a trapezoid the mid-line =the average between the big base and small base ½(big+small)
Congruent Cevians Theorems:
1. corresponding medians of congruent triangles are congruent
triangle ABC is congruent to triangle DEF  given
A
AB is congruent to ED  CPCTC
<<A is congruent to <<D CPCTC
AR is congruent to RC definition of median
DS is congruent to SF definition of median
AR is congruent to DS  subtraction
Triangle ABR is congruent to triangle DES  SAS
BR is congruent to ES CPCTC
D
R
S
B
E
C
C
F