Ch 4 Notesheet Key
Ch 4 Notesheet Key
Ch 4 Notesheet Key
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
<strong>Ch</strong>apter 4:<br />
Discovering and Proving<br />
Triangle Properties<br />
Note Sheet<br />
S. Stirling Page 1 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
4.1 Triangles Sum Conjectures<br />
These notes replace pages 200 – 202 in the book. See step 4 of the Investigation.<br />
Rigidity is a property that triangles have. They cannot be shifted, like a quadrilateral can. They retain<br />
their shape.<br />
Triangulation is a procedure used by surveyors to locate a position by using triangles.<br />
Triangle Sum Conjecture<br />
The sum of the measures of the angles in every triangle is 180°.<br />
Numeric Example:<br />
Find x.<br />
A<br />
56<br />
B<br />
101<br />
101 + 56 + x = 180<br />
x = 180 – 157 = 23<br />
x<br />
C<br />
Find y.<br />
x + x + y = 180<br />
2x + y = 180<br />
y = 180 – 2x<br />
E<br />
x<br />
F<br />
x<br />
y<br />
D<br />
Third Angle Conjecture<br />
If two angles of one triangle are congruent to two angles of another triangle, then the third<br />
angles of the triangles are congruent.<br />
Numeric Example:<br />
If m∠ K = m∠Band m∠ I = m∠ C , find y. Give reasons for your answer!<br />
K<br />
101<br />
B<br />
A<br />
y<br />
56 23<br />
J<br />
I<br />
C<br />
y = 56 because the third angles are equal also.<br />
See <strong>Ch</strong> 4 Worksheet to prove these conjectures.<br />
S. Stirling Page 2 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
Lesson 4.2 Properties of Isosceles Triangles<br />
Read top of page 206. This vocabulary is review! Know it!!<br />
Remember! With If..then.. statements: If a, then b. The converse of the statement is If b, then a.<br />
Label and define the special vocabulary for the isosceles triangle Δ VBS .<br />
B<br />
A leg (of an isosceles triangle) is one of the congruent<br />
sides. VB and VS<br />
S<br />
The base (of an isosceles triangle) is the side that is not a<br />
leg. BS<br />
The vertex angle is the angle between the two legs.<br />
∠ V<br />
V<br />
The base angles are the pair of angles whose vertices are<br />
the endpoints of the base. ∠B<br />
and ∠ S<br />
Do group Investigation 1, Base angles of an Isosceles Triangle on page 207.<br />
Isosceles Triangle Conjecture<br />
If a triangle is isosceles, then its base angles are congruent.<br />
Given<br />
If VB<br />
ΔVBS<br />
with VB = VS , then ∠B ≅∠ S or<br />
≅ VS , then ∠B ≅ ∠ S .<br />
B<br />
V<br />
S<br />
***Note: Since equilateral triangles are a special case of isosceles triangles, any property<br />
that applies to isosceles triangles also apply to equilateral triangles.<br />
Do group Investigation 2, Is the Converse True? on page 208. Use a protractor.<br />
Converse of the Isosceles Triangle Conjecture<br />
If a triangle has two congruent angles, then it is an isosceles triangle.<br />
If ∠B ≅ ∠ S , then ΔVBS<br />
is isosceles or<br />
If ∠B ≅ ∠ S , then VB ≅ VS .<br />
V<br />
B<br />
S<br />
S. Stirling Page 3 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Lesson 4.3 Triangle Inequalities<br />
Read of page 215. Do group Investigation 1 on page 216.<br />
Name ___________________________<br />
Triangle Inequality Conjecture<br />
The sum of the lengths of any two sides of a triangle is greater than<br />
third side.<br />
Is a triangle measuring 5 cm, 4 cm and 2 cm<br />
possible? Construct it then explain!<br />
the length of the<br />
Is a triangle measuring 5 cm, 3 cm and 2 cm<br />
possible? Construct it then explain!<br />
Since 2 + 4 > 5, it makes a triangle.<br />
Since 2 + 3 = 5, it makes segment NOT<br />
a triangle.<br />
Do group Investigation 2, Largest and Smallest Angles in a Triangle? on page 217.<br />
Side-Angle Inequality Conjecture (Only applies to triangles!)<br />
In a triangle, if one side is the longest side, then the angle opposite the longest side is the<br />
largest angle.<br />
Likewise, if one side is the shortest side, then the angle opposite the shortest side is the<br />
smallest angle.<br />
Which is the largest angle? The smallest<br />
angle? Why?<br />
K<br />
J<br />
2.48<br />
6.3<br />
5.34<br />
I<br />
Which is the largest side? Why?<br />
∠D<br />
largest, so AQ<br />
largest in Δ ADQ ,<br />
But, m∠ QAU = 100<br />
D<br />
A<br />
90<br />
54<br />
23<br />
Q<br />
∠ K largest, its opposite the longest<br />
side, JI .<br />
∠I<br />
smallest, its opposite the shortest<br />
side, JK .<br />
which makes UQ the<br />
largest in ΔQAU<br />
and<br />
AQ < UQ.<br />
Overall, UQ is the longest.<br />
57<br />
U<br />
S. Stirling Page 4 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
Do group Investigation 3, Exterior Angle of a Triangle on page 217 – 218 .<br />
Label the drawing with the following terms and define the terms:<br />
Exterior angle is an angle that forms a linear pair<br />
with one of the interior angles of a polygon.<br />
Adjacent interior angle is the angle of a polygon<br />
that forms a linear pair with a given exterior angle of<br />
a polygon.<br />
The remote interior angles (of a triangle) are the<br />
interior angles of a triangle that do not share a vertex<br />
with a given exterior angle.<br />
Triangle Exterior Angle Conjecture<br />
The measure of an exterior angle of a triangle is equal to the sum of the measures of the<br />
remote interior angles.<br />
Given ΔABC<br />
above. If a = 50 and b = 60, what<br />
is the measure of ∠ BCD ? Explain.<br />
50 + 60 = 110<br />
The exterior angle equals the sum of the<br />
two remote interior angles.<br />
Given ΔABC<br />
above. If x = 80 and b = 30, what<br />
is the measure of ∠ A ? Explain.<br />
80 – 30 = 50<br />
The exterior angle equals the sum of the<br />
two remote interior angles.<br />
Before starting Lesson 4.4:<br />
Vocab for triangles Triangles have 6 “parts”.<br />
∠A<br />
is opposite BC and AC is opposite ∠ B<br />
∠A<br />
is between BA and AC (angles are between sides)<br />
BC is between<br />
∠ B and ∠ C (sides are between angles)<br />
included angle is an angle formed between two consecutive<br />
sides of a polygon.<br />
B<br />
A<br />
C<br />
included side is a side of a polygon between two consecutive angles.<br />
S. Stirling Page 5 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
Read the top of page 221, then complete the Triangle Congruence Shortcut Investigation, page 1 – 3.<br />
SSS Congruence Conjecture<br />
If the three sides of one triangle are congruent to the three sides of another triangle, then<br />
the triangles are congruent.<br />
A<br />
D<br />
O<br />
A<br />
G<br />
C<br />
T<br />
ΔCAT<br />
≅ Δ _______<br />
By SSS Congruence Conjecture.<br />
C<br />
T<br />
O<br />
ΔCAT<br />
≅Δ _______<br />
By SSS Congruence Conjecture.<br />
SAS Congruence Conjecture<br />
If two sides and the included angle of one triangle are congruent to two sides and the<br />
included angle of another triangle, then the triangles are congruent.<br />
A<br />
I<br />
P<br />
I<br />
Z<br />
C<br />
T Z P<br />
ΔPIZ<br />
≅ Δ _______<br />
By SAS Congruence Conjecture.<br />
E<br />
D<br />
ΔZED<br />
≅Δ _______<br />
By SAS Congruence Conjecture.<br />
ASA Congruence Conjecture<br />
If two angles and the included side of one triangle are congruent to two angles and the<br />
included side of another triangle, then the triangles are congruent.<br />
D<br />
O<br />
I<br />
O<br />
G<br />
Z<br />
P<br />
ΔDOG<br />
≅ Δ _______<br />
By ASA Congruence Conjecture.<br />
C<br />
P<br />
A<br />
ΔCOP<br />
≅Δ _______<br />
By ASA Congruence Conjecture.<br />
S. Stirling Page 6 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
SAA or AAS Congruence Conjecture<br />
If two angles and a non-included side of one triangle are congruent to the corresponding<br />
angles and side of another triangle, then the triangles are congruent.<br />
D<br />
O I<br />
G<br />
Z<br />
P<br />
ΔPIZ<br />
≅ Δ _______<br />
By SAS Congruence Conjecture.<br />
ΔCAP<br />
≅Δ _______<br />
P<br />
A<br />
T<br />
C<br />
By SAS Congruence Conjecture.<br />
Special Case: Hypotenuse Leg Congruence Conjecture<br />
If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one<br />
leg of another right triangle, then the two triangles are congruent.<br />
D<br />
O<br />
I<br />
O<br />
G<br />
Z<br />
P<br />
ΔDOG<br />
≅ Δ _______<br />
By HL Congruence Conjecture.<br />
C<br />
P<br />
A<br />
ΔCOP<br />
≅Δ _______<br />
By HL Congruence Conjecture.<br />
SSA or ASS Congruence?<br />
If two sides and the non-included angle of one triangle are congruent to two sides and the<br />
non-included angle of another triangle, then the triangles are NOT necessarily congruent.<br />
A<br />
I<br />
Draw a counterexample.<br />
I<br />
C<br />
T Z<br />
P<br />
ΔZIP<br />
≅ Δ _______<br />
By NOT necessarily congruent.<br />
Z<br />
P<br />
S. Stirling Page 7 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
AAA Congruence Conjecture<br />
If three angles of one triangle are congruent to the corresponding angles of another<br />
triangle, then the triangles are NOT necessarily congruent.<br />
A<br />
I<br />
Draw a counterexample.<br />
I<br />
C<br />
T Z<br />
P<br />
ΔZIP<br />
≅ Δ _______<br />
By NOT necessarily congruent.<br />
Z<br />
P<br />
How do you get equal parts in order to get congruent triangles?<br />
To see if you have congruent triangles, you will be checking for a SSS, SAS, ASA or AAS marked on the matching pair of<br />
triangles that you be given (as shown above). Sometimes the parts will be marked equal in the diagram. That’s the easy stuff.<br />
Other times, you will be given information that you will “translate” into equal sides and angles in order to get your<br />
congruence. You will need to deduce this information from definitions or conjectures that you already know to be true.<br />
Complete the following to help you review these statements. Remember, to mark your diagrams with the equal parts. Also<br />
never assume things are congruent! You must have a definitions or conjecture to back you up!!<br />
Ways to Get Equal Segments<br />
Converse of the Isosceles Triangle Conj.<br />
If a triangle has two congruent angles,<br />
then it is an isosceles triangle. V S<br />
Def. Isosceles Triangle<br />
If a triangle is isosceles,<br />
then its legs are congruent.<br />
V<br />
S<br />
If ∠B ≅ ∠ S ,<br />
≅ VS<br />
then VB<br />
B<br />
ΔVBS<br />
If isosceles<br />
then VB ≅ VS .<br />
B<br />
Def. Midpoint<br />
If a point is a midpoint,<br />
then it divides the segment into<br />
two equal segments.<br />
If M is the midpoint of SG ,<br />
then SM ≅ MG .<br />
S<br />
M<br />
G<br />
Def. segment bisector<br />
If a line (or part of a line) is a bisector,<br />
then it passes through the<br />
midpoint of the segment.<br />
If CM bisects AB ,<br />
≅ MB .<br />
then AM<br />
A<br />
C<br />
M<br />
B<br />
“Same Segment”<br />
When two triangles share<br />
the exact same segment, you get<br />
a pair of equal segments.<br />
ΔBET<br />
and ΔWTE<br />
ET ≅ TE<br />
share<br />
B<br />
E<br />
O<br />
T<br />
W<br />
Def. Median<br />
If a segment is a median,<br />
then it connects the vertex to the<br />
midpoint of the opposite side.<br />
If AM is median in Δ ABC ,<br />
then BM ≅ MC .<br />
A<br />
B<br />
M<br />
C<br />
S. Stirling Page 8 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Isosceles Triangle Conjecture<br />
If a triangle is isosceles,<br />
then its base angles are congruent.<br />
ΔVBS<br />
If isosceles<br />
or VB ≅ VS ,<br />
then ∠B ≅ ∠ S .<br />
Definition of Angle Bisector<br />
If you have an angle bisector,<br />
then the ray cuts the angle<br />
into two equal angles.<br />
<br />
If BD bisects ∠ ABC<br />
then ∠ABD ≅ ∠ DBC .<br />
Def. Altitude<br />
If a segment is an altitude,<br />
then it goes from a vertex<br />
perpendicular to the line that<br />
contains the opposite side.<br />
Δ TRI<br />
∠ = ∠ = °.<br />
If RA is an altitude of<br />
then m RAT m RAI 90<br />
“Same Angle”<br />
When two triangles share<br />
the exact same angle, you get<br />
a pair of equal angles.<br />
ΔTIA<br />
and ΔNIR<br />
∠TIA<br />
≅ ∠ RIN .<br />
T<br />
share<br />
T<br />
B<br />
V<br />
B<br />
R<br />
Name ___________________________<br />
Ways to Get Equal Angles<br />
I<br />
G<br />
R<br />
A<br />
S<br />
C<br />
A<br />
A<br />
D<br />
I<br />
N<br />
Vertical Angle Conjecture<br />
If two angles are vertical,<br />
then they are congruent.<br />
If ∠VET<br />
and ∠CER<br />
then ∠VET ≅∠ CER .<br />
Def. perpendicular lines<br />
If two lines are perpendicular,<br />
then they intersect to form equal<br />
90° angles.<br />
If CD<br />
V<br />
C<br />
E<br />
are vertical,<br />
⊥ AB , then<br />
m∠ CMA= m∠ CMB = 90°.<br />
Corresponding Angles Conjecture<br />
If two parallel lines are cut by a transversal,<br />
then corresponding angles are congruent.<br />
<br />
<br />
If AB DC ,<br />
then ∠EFA ≅∠ FGD .<br />
Alternate Interior Angles Conjecture<br />
If two parallel lines are cut by a transversal,<br />
then alternate interior angles are congruent.<br />
<br />
If LK NJ ,<br />
then ∠KHM ≅∠ HMN .<br />
A<br />
D<br />
C<br />
N<br />
A<br />
L<br />
M<br />
G<br />
E<br />
M<br />
T<br />
F<br />
H<br />
R<br />
B<br />
D<br />
B<br />
C<br />
K<br />
J<br />
Ways to Get equal Angles and Sides<br />
Complete the Triangle<br />
Congruence Shortcut<br />
Investigation, page 4 – 5.<br />
Def. perpendicular bisector<br />
A line (or part of a line) that passes through<br />
the midpoint of a segment and<br />
is perpendicular to the segment.<br />
If CD is the perp. bisector of AB ,<br />
then AM ≅ MB<br />
∠ CMA =∠ CMB = °.<br />
and 90<br />
C<br />
A<br />
M<br />
B<br />
D<br />
S. Stirling Page 9 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
4.6 Corresponding Parts of Congruent Triangles<br />
Note: Once you can prove that two triangles are congruent, by using the conjectures above, then any of<br />
the corresponding parts will be equal.<br />
Definition of Congruent Triangles:<br />
If two triangles are congruent, then all of their corresponding parts (sides and angles) are<br />
congruent.<br />
Can also be stated “Corresponding Parts of Congruent Triangles are Congruent” or<br />
CPCTC.<br />
When you write congruent triangles you must match the corresponding vertices. (To show how the match up.)<br />
So if you know that ΔCAT<br />
≅ ΔDOG<br />
then you can say any of the following:<br />
∠C<br />
≅ ∠ D CA ≅ DO<br />
∠A<br />
≅ ∠ O AT ≅ OG<br />
∠T<br />
≅ ∠ G CT ≅ DG<br />
Examples: How to prove (show deductively) that parts are equal.<br />
First example, informal paragraph :<br />
P<br />
Know: ∠PAC ≅ ∠TAC<br />
and PA AT<br />
Is ∠PCA ≅ ∠ TCA ?<br />
≅ .<br />
A<br />
T<br />
The triangles share a side, so CA<br />
So…<br />
= CA of course.<br />
C<br />
ΔPAC<br />
≅ ΔTAC<br />
by SAS Congruence<br />
and ∠PCA ≅ ∠ TCA because CPCTC or Corresponding Parts of Congruent Triangles are<br />
Congruent.<br />
A bit more formal paragraph:<br />
Given: Z is midpoint of DI and EP .<br />
Z<br />
Prove: PI ≅ DE ?<br />
Z is midpoint of DI and EP was given, so<br />
IZ = ZD and PZ = ZE by Definition of Midpoint<br />
D<br />
E<br />
∠PZI<br />
≅ ∠ EZD by the Vertical Angles Conjecture (or vertical angles are congruent.)<br />
Therefore, ΔZIP ≅ ΔZDE<br />
by SAS Congruence and<br />
PI DE<br />
≅ because CPCTC or Corresponding Parts of Congruent Triangles are Congruent.<br />
P<br />
I<br />
S. Stirling Page 10 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
This one is tricky:<br />
Know: RE<br />
Is RE ≅<br />
EC<br />
TC and TR CE .<br />
? What is congruent?<br />
Name ___________________________<br />
RE TC given so ∠RET ≅ ∠ETC<br />
and<br />
TR CE given so ∠RTE ≅ ∠CET<br />
because parallel lines make equal alternate interior angles.<br />
TE = TE the triangles share a side,<br />
ΔTRE<br />
≅ ΔECT<br />
by ASA Congruence. So any of the corresponding sides are congruent.<br />
RE and EC are NOT corresponding sides. So, probably not equal.<br />
Note: you could say any of the following though: RE<br />
T<br />
R<br />
C<br />
≅ TC , RT ≅ EC or ∠R ≅ ∠ C .<br />
E<br />
This one is trickier! Start by un-overlapping possible pairs of congruent triangles:<br />
Given: LE = EP and LV = OP<br />
Prove: ∠ALP ≅ ∠ APL<br />
O<br />
A<br />
V<br />
E<br />
L<br />
P<br />
LE = EP so ∠ELP ≅ ∠ EPL If isosceles, then base angles are equal.<br />
LP = PL Same segment<br />
LV = OP Given information<br />
∠LOP<br />
≅ ∠PVL<br />
by SAS Congruence<br />
∠ALP<br />
≅ ∠ APL CPCTC<br />
Try Example A and Example B on <strong>Ch</strong> 4 Worksheet.<br />
S. Stirling Page 11 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
4.7 Flowchart Thinking<br />
4.7 Page 238 Example B Flowchart Proof<br />
Given: EC AC<br />
≅ and ER ≅ AR<br />
Prove: ∠A ≅ ∠ E<br />
Name ___________________________<br />
E<br />
R<br />
C<br />
EC<br />
Given<br />
≅<br />
AC<br />
A<br />
ER ≅ AR<br />
Given<br />
RC≅<br />
RC<br />
Same segment<br />
ΔREC<br />
≅ ΔRAC<br />
SSS Cong. Conj.<br />
∠A<br />
≅∠E<br />
CPCTC<br />
4.7 Page 239 Top. Explain why the angle bisector construction works. Flowchart Proof<br />
Given:<br />
∠ABC<br />
with BA ≅ BC and CD ≅ AD<br />
ABC<br />
Prove: BD is the angle bisector of<br />
∠ .<br />
BA ≅<br />
BC<br />
Given<br />
CD ≅ AD<br />
Given<br />
BD≅<br />
BD<br />
Same segment<br />
ΔBAD<br />
≅ ΔBCD<br />
SSS Cong. Conj.<br />
∠1≅∠2<br />
CPCTC<br />
BD is the angle bisector of<br />
∠ ABC .<br />
Def. of angle bisector<br />
S. Stirling Page 12 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
What do you need to know to get…<br />
You’ll need Equal Segments to get…<br />
Def. Isosceles Triangle<br />
If a triangle’s legs are congruent,<br />
then it is isosceles.<br />
If VB<br />
then<br />
≅ VS<br />
ΔVBS<br />
isosceles.<br />
B<br />
V<br />
S<br />
Def. Median<br />
If a segment connects the vertex to the<br />
midpoint of the opposite side,<br />
then it is a median.<br />
If BM ≅ MC ,<br />
then AM is median in<br />
A<br />
Δ ABC .<br />
B<br />
M<br />
C<br />
Def. Midpoint<br />
S<br />
If a point divides the segment into<br />
two equal segments,<br />
then it is a midpoint.<br />
If SM ≅ MG ,<br />
then M is the midpoint of SG .<br />
M<br />
G<br />
Def. segment bisector<br />
C<br />
If a line (or part of a line) passes through the<br />
midpoint of the segment,<br />
then it is a bisector.<br />
M<br />
If AM ≅ MB,<br />
A<br />
then CM bisects AB .<br />
B<br />
You’ll need Equal Angles to get…<br />
Converse of the Isosceles Triangle Conj.<br />
If a triangle has two congruent angles,<br />
then it is an isosceles triangle. V S<br />
If ∠B ≅ ∠ S ,<br />
ΔVBS<br />
then<br />
isosceles.<br />
B<br />
Def. perpendicular lines<br />
If two lines intersect to form equal<br />
adjacent angles,<br />
then they are perpendicular.<br />
If m∠ CMA= m∠ CMB ,<br />
⊥ AB .<br />
then CD<br />
C<br />
A<br />
M<br />
B<br />
D<br />
Definition of Angle Bisector<br />
If a ray cuts the angle into two<br />
equal angles, then you have<br />
an angle bisector.<br />
If ∠ABD ≅ ∠ DBC ,<br />
ABC<br />
then BD bisects ∠ .<br />
B<br />
C<br />
A<br />
D<br />
Corresponding Angles Conjecture<br />
If two lines are cut by a transversal and<br />
the corresponding angles are congruent,<br />
then the lines are parallel.<br />
If ∠EFA ≅∠ FGD ,<br />
<br />
DC<br />
then AB .<br />
A<br />
D<br />
G<br />
E<br />
F<br />
B<br />
C<br />
Def. Altitude<br />
If a segment goes from a vertex<br />
perpendicular to the line that<br />
contains the opposite side,<br />
then it is an altitude.<br />
If m∠ RAT = m∠ RAI = 90°,<br />
TRI<br />
then RA is an altitude of<br />
T<br />
R<br />
A<br />
Δ .<br />
I<br />
Alternate Interior Angles Conjecture<br />
If two lines are cut by a transversal and<br />
the alternate interior angles are congruent,<br />
then the lines are parallel.<br />
If ∠KHM ≅∠ HMN ,<br />
<br />
NJ<br />
then LK<br />
.<br />
L<br />
N<br />
M<br />
H<br />
K<br />
J<br />
S. Stirling Page 13 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
Name ___________________________<br />
You’ll need Equal Angles and Sides to get…<br />
Def. perpendicular bisector<br />
If a line (or part of a line) that passes through the midpoint of<br />
a segment and is perpendicular to the segment,<br />
then it is the perpendicular bisector.<br />
C<br />
If AM<br />
≅ MB and ∠ CMA = 90°,<br />
then CD is the perp. bisector of AB .<br />
A<br />
M<br />
B<br />
D<br />
Summary<br />
Basic Procedure for Proofs<br />
“parts” refers to sides and/or angles.<br />
1. Get equal parts by using given info. and known definitions and conjectures.<br />
2. State the triangles are congruent by SSS, SAS, ASA or AAS.<br />
3. Use CPCTC to get more equal parts.<br />
4. Connect that info. to what you were trying to prove .<br />
Hints [if you get stuck]:<br />
Mark the diagram with what you have stated as congruent in your proof.<br />
( If given M is the midpoint of AB , convert it to AM = MB by def. of<br />
midpoint before marking the diagram!)<br />
Look at the diagram to find equal parts.<br />
Brainstorm and then apply previous conjectures and definitions.<br />
Work (or think) backwards!<br />
Draw overlapping triangles separately.<br />
Re-draw figures without all of the “extra segments” in there.<br />
Draw an auxillary line.<br />
Break a problem into smaller parts.<br />
S. Stirling Page 14 of 15
<strong>Ch</strong> 4 Note Sheet L1 <strong>Key</strong><br />
4.8 Proving Special Triangle Conjectures<br />
Name ___________________________<br />
Equilateral/Equiangular Triangle Conjecture<br />
Every equilateral triangle is equiangular.<br />
Conversely, every equiangular triangle is equilateral.<br />
Vertex Angle Bisector Conjecture<br />
In an isosceles triangle, the bisector of the vertex angle is also the altitude and the median<br />
to the base and the perpendicular bisector of the base.<br />
Isosceles Triangle with vertex angle A, AB = AC.<br />
Median BM<br />
A<br />
Altitude BL<br />
L<br />
Angle Bisector BS<br />
AB = 3.36 in.<br />
AC = 3.36 in.<br />
CB = 4.34 in.<br />
C<br />
M<br />
S<br />
B<br />
Isosceles Triangle with Vertex Angle B<br />
B<br />
A<br />
EMD<br />
C<br />
Median BM<br />
AM = 1.64 in.<br />
MC = 1.64 in.<br />
Altitude BE<br />
m∠AEB = 90°<br />
Angle Bistector BD<br />
m∠ABD = 27°<br />
m∠DBC = 27°<br />
Medians to the Congruent Sides Theorem<br />
In an isosceles triangle, the medians to the congruent sides are congruent.<br />
Angle Bisectors to the Congruent Sides Theorem<br />
In an isosceles triangle, the angle bisectors to the congruent sides are congruent.<br />
Altitudes to the Congruent Sides Theorem<br />
In an isosceles triangle, the altitudes to the congruent sides are congruent.<br />
A<br />
A<br />
A<br />
N<br />
M<br />
S<br />
P<br />
T<br />
L<br />
C<br />
C<br />
C<br />
B<br />
B<br />
B<br />
Medians MC = BN Angle Bisectors PC = BS Altitudes BT = LC<br />
S. Stirling Page 15 of 15