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Geometry - Semester 2
Mrs. Day-Blattner
2/22/2016
Agenda 2/22/2016
1. Turn in homework - Arcs of a Circle Table and Geometry:
Semester 1 Review cont.
2. Lesson 8 Puzzles involving circles and angles (p. 53-60).
a. Part 1 Pair - then share
b. Part 2
3. Problem Set
4. Exit ticket
Part 1: Give a reason for each step of this proof. Add
missing measurements as you go.
In a circle, a chord DE
and a diameter AB are
extended outside of
the circle to meet a A
point C.
D
46°
x°
.
O
If m∠DAE = 46° and
m∠DCA = 32° find m∠DEA = 32°
y°
E
32°
B
C
Let m∠DEA = y and m∠EAB = x
D
In triangle ABD,
m∠DBA = y
46°
A
x°
.
O
Reason?
Think-pair each step
3 mins
y°
E
y°
32°
B
C
Let m∠DEA = y and m∠EAB = x
D
In triangle ABD,
m∠DBA = y
46°
A
x°
.
O
Reason: Angles
inscribed in same arc
are congruent.
y°
E
y°
32°
B
C
m∠ADB = 90°
D
Reason?
46°
A
x°
.
O
y°
E
32°
y°
B
C
m∠ADB = 90°
D
90°
Reason: For a triangle
x°
A
inscribed in a circle if one
side is a diameter it is the
hypotenuse of a right
triangle. Thales’ theorem.
46°
.
O
y°
E
y°
32°
B
C
therefore 46 + x + y + 90 = ?
Reason?
D
90°
46°
A
x°
.
O
y°
E
32°
y°
B
C
therefore 46 + x + y + 90 = 180
x + y = 180 - 46 - 90
x +y = 44
Reason: Sum of
angles of a triangle
is 180 degrees.
D
90°
46°
A
x°
.
O
y°
E
32°
y°
B
C
In triangle ACE y = x + 32
D
90°
Reason?
46°
A
x°
.
O
y°
E
32°
y°
B
C
In triangle ACE y = x + 32
D
90°
Reason: External
angle of a triangle
is equal to the sum
of the remote
interior angles
46°
A
x°
.
O
y°
E
32°
y°
B
C
y = x + 32
x + 32 = y
x + x + 32 = x + y
D
90°
Reason: Addition
property of
equality
46°
A
x°
.
O
y°
E
32°
y°
B
C
y = x + 32
x + 32 = y
x + x + 32 = x + y
2x + 32 = 44
D
90°
Reason:
Substitution
property of
equality
46°
A
x°
.
O
y°
E
32°
y°
B
C
y = x + 32
x + 32 = y
x + x + 32 = x + y
2x + 32 = 44
D
90°
46°
A
What next?
x°
.
O
y°
E
32°
y°
B
C
y = x + 32
x + 32 = y
x + x + 32 = x + y
2x + 32 = 44
D
90°
2x = 44 - 32
x = 12/2
x=6
46°
A
x°
.
O
y°
E
32°
y°
B
C
y = x + 32
x + 32 = y
x + x + 32 = x + y
2x + 32 = 44
D
90°
2x = 44 - 32
x = 12/2
x=6
6 + y = 44
y = 44 - 6
y = 38
46°
A
x°
.
O
y°
E
32°
y°
B
C
y = x + 32
x + 32 = y
x + x + 32 = x + y
2x + 32 = 44
D
90°
2x = 44 - 32
x = 12/2
x=6
6 + y = 44
y = 44 - 6
y = 38
46°
A
x°
.
O
m∠DEA = 38°
y°
E
32°
y°
B
C
Part 2: Solve for x. Give a reason for every step.
1. Hint: Thales’ theorem “ “
2. Central angle twice inscribed angle for same intercepted
arc.
3. Same as 2.
4. Draw a center of circle A.
Problem Set.
1, 2, 4, and 6, 7a. (7b. extra credit challenge)
Exit - ticket - turn in before leave - Collect homework sheet
Part 2: Solve for x. Give a reason for every step.
1. Hint: Thales’ theorem
We have two triangles each with one side being the diameter
of the circle. Hence, by Thales’ theorem, we have 2 right
triangles inscribed in the circle.
Angle BEC and angle BDC are both 90 degrees.
Also Triangle BCE is an isosceles triangle, since the legs BE
and EC are marked as congruent. For an isosceles triangle the
base angles are also congruent, and since the sum of the
meausures of the 3 angles in a triangle will be 180 degrees we
can find the base angles of triangle BCe - they are both 45
degrees. (next slide)
Part 2: Solve for x. Give a reason for every step.
1. cont.
Since angle EBC which is 45 degrees, intercepts the same arc
(arc EC) as angle CDE they must be congruent (inscribed
angles intercepting the same arc are congruent).
measure of Angle CDE is 45 degrees
hence x = 45
Part 2: Solve for x. Give a reason for every step.
2. measure of angle CBE is 34 degrees (given).
Since for parallel lines cut by a transversal, corresponding
angles are congruent, we know that the measure of angle
CAD must also be 34 degrees, since angle CBE and angle
CAD are corresponding angles relative to the two parallel line
segments (AD and BE).
angle CAD and angle DAB are a linear pair of angles, are
supplementary
hence measure of angle CAD + measure of angle DAB = 180
degrees
measure of angle DAB = 180 degrees - 34 deg. = 146
degrees.
Part 2: Solve for x. Give a reason for every step.
2. cont.
measure of angle DAB = 180 degrees - 34 deg. = 146
degrees.
angle DAB is a central angle that intercepts arc BE.
measure of angle ADE = ½ measure DAB by inscribed angle
theorem, since both angles intercept the same arc.
measure of angle ADE = ½(146 deg) = 73 degrees.
x = 73
Part 2: Solve for x. Give a reason for every step.
3. measure of angle CFB = measure of angle BEC
because angles that are inscribed in a circle and intercept the
same arc are congruent.
The inscribed angles intercept the same arc as central angle
BAC marked as 104 deg. and they will be half the meausure of
the central angle, using the inscribed angle theorem.
Hence measure Angle CFB = measure of angle BEC = ½ (104
deg) = 52 degrees.
Angles DEG and BEC are linear pairs, so their sum is 180
degrees, so measure of angle DEG = 180 - 52 = 128 deg.
and measure of angle GFD will be the same.
Part 2: Solve for x. Give a reason for every step.
3. cont.
Now GEDF is a quadrilateral whose internal angles will add up
to 180 degrees.
The measure of angle EGF = (360 - 30 - 128 - 128) degrees
m. angle EDF = 74 degrees.
x = 74
since the angle ECF and BGC are vertical angles and hence
congruent.
Part 2: Solve for x. Give a reason for every step.
4. Draw a center A.
The measure of angle EAF will be 60 degrees since it is a
central angle that intercepts the same arc, EF, as the inscribed
angle ECF marked as 30 degrees. (The central angle is twice
the measure of an inscribed angle intercepting the same arc,
this is the inscribed angle theorem.)
If we mark in two congruent radii of the circle we can see that
triangle AEF is an isosceles triangle and so it will have
congruent base angles and since the sum of the measures of
the angles in a triangle are 180 degrees we can show that the
base angles are both 60 degrees.
Part 2: Solve for x. Give a reason for every step.
4. cont.
We can draw 2 more isosceles triangles now and see that the
line segment DC through A is a diameter of the circle, which is
a central angle of 180 degrees.
Inscribed angle DBC intercepts the same arc (DEFC) as the
diameter and so by the inscribed angle theorem, angle DBC
will be half the measure of the central angle DC,
½ (180 deg.) = 90 deg
Hence, x = 90
(or angle inscribed in a semicirle = 90 degrees).
Problem set.
1. Find the value x.
angle BAD is a central angle (vertex at center of circle)
angle BED is an inscribed angle (vertex on the circle) that
intercepts the same arc, BD.
By inscribed angle theorem
x = ½ measure of angle BAD
x = ½ 81 = 40.5
Problem set.
2. Find value of x.
Notice that inscribed angle BED (57 degrees) intercepts the
arc BD and that inscribed angle BCD intercepts the same arc.
As a consequence of the inscribed angle theorem, we know
that inscribed angles that intercept the same arc are equal in
measure.
Hence x = 57
Problem set.
4. Find value of x.
measure angle EBC = 124 degrees, given
measure angle ABE = 180 degrees - 124 degrees = 56 deg
since they are linear pairs (supplementary angles)
triangle ABE is an isosceles triangle since it has 2 legs that are
congruent (radii of the circle)
Measure of angle AEB = 56 deg
measure of angle EAB (central angle) = 180 - 56 -56 = 68 deg.
x is inscribed angle that intercepts same arc (EB) as the
central angle with measure 68 deg, hence x = ½ 68 = 34
Problem set.
6. If BF = FC, express y in terms of x.
Inscribed angle EDC = y and it intercepts arc EF.
Inscribed angle EBF intercepts the same arc (EF) and so as a
consequence of the inscribed angle theorem, angles EDC and
EBF must be congruent.
If BF = FC then BFC is an isosceles triangle with 2 congruent
base angles, we already labelled angle EBF with y, so angle
BCF is also y.
Now sum of angles of a triangle = 180
for triangle DEC 180 = y + y + (180 - x)
or x = 2y
Problem set.
7.a. Find the value x
Line segment AD is parallel to line segment CB and is cut by
transversal line segment CD.
Angles ECB and EDA are congruent since they are alternate
interior angles.
Inscribed angle DCB intercepts the same arc as central angle
BAD, so the central angle is 2(inscribed angle) by the
inscribed angle theorem.
measure of angle BAD = 2 (30 degrees) = 60 degrees
Problem set.
7.a. Find the value x
measure of angle BAD = 2 (30 degrees) = 60 degrees
For triangle AED we know two of its angles are 60 degrees
and 30 degrees and so the third angle must be 90 degrees
since the sum of the angles of a triangle is 180 degrees.
measure of Angle BED + measure of angle AED = 180
degrees since they are supplementary /linear pairs.
Hence x = 90
Lesson Summary
Inscribed angle theorem: The measure of a central angle is
twice the measure of an inscribed angle that intercepts the
same arc as the central angle.Consequence of inscribed
angle theorem: Inscribed angles that intercept the same arc
are equal in measure.
If A, B, B’, and C are four points with B and B’ on
the same side of line AC, and angles ABC and
AB’C are congruent, then A, B, B’ and C all lie on
the same circle.
Homework - due on Wednesday
5 sample questions for quiz on Friday.
1- 4 you solve and show how each time (state
theorem itself not just its name).
5. Create your own problem. Give question and
its solution.