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TEKS Focus:
(6)(B) Prove two triangles are congruent by applying the Side-Angle-Side,
Angle-Side-Angle, Side-Side-Side, Angle-Angle-Side, and Hypotenuse-Leg
congruence conditions.
(1)(C) Select tools, including real objects, manipulatives paper and pencil,
and technology as appropriate, and techniques, including mental math,
estimations, and number sense as appropriate, to solve problems.
(1)(E) Create and use representations to organize, record, and communicate
mathematical ideas.
(1)(F) Analyze mathematical relationships to connect and communicate
mathematical ideas.
(1)(G) Display, explain, or justify mathematical ideas and arguments using
precise mathematical language in written or oral communication.
(5)(A) Investigate patterns to make conjectures about geometric
relationships, including angles formed by parallel lines cut by a transversal,
criteria required for triangle congruence, special segments of triangles,
diagonals of quadrilaterals, interior and exterior angles of polygons, and
special segments and angles of circles choosing from a variety of tools.
(5)(C) Use the constructions of congruent segments, congruent angles, angle
bisectors, and perpendicular bisectors to make conjectures about geometric
relationships.
An included angle is an angle formed by
two adjacent sides of a polygon.
B is the included angle between sides AB
and BC.
The property of triangle rigidity
gives you a shortcut for proving two
triangles congruent. It states that if
the side lengths of a triangle are
given, the triangle can have only
one shape.
For example, you only need to know that
two triangles have three pairs of congruent
corresponding sides. This can be expressed
as the following postulate.
It can also be shown that only two pairs of
congruent corresponding sides are needed
to prove the congruence of two triangles if
the included angles are also congruent.
Caution
The letters SAS are written in that order
because the congruent angles must be
between pairs of congruent corresponding
sides.
Remember!
Adjacent triangles share a side, so you can apply the
Reflexive Property to get a pair of congruent parts.
EXAMPLE 1:
The triangles can be proved congruent by SSS.
They cannot be proven congruent by SAS because
the angles are in different positions.
LB is congruent to itself by the Reflexive Property.
That gives us 2 out of the 3 parts.
We must have the side that makes ELB and NBL the
included angle.
Therefore, we need LE  NB.
Use SSS to explain why ∆ABC  ∆DBC.
Statements
Reasons
1. AC  DC
1. Given
2. AB  DB
2. Given
3. BC  BC
3. Reflexive Property of
Congruence
4. ABC  DBC
4. SSS
Use SSS to explain why
∆ABC  ∆CDA.
Statements
Reasons
1. AB  CD
1. Given
2. BC  DA
2. Given
3. AC  AC
3. Reflexive Property of
Congruence
4. ABC  CDA
4. SSS
The diagram shows part of the
support structure for a tower. Use
SAS to explain why ∆XYZ  ∆VWZ.
Statements
Reasons
1. Given
1. XZ  VZ
2.
XZY 
VZW
3. YZ  WZ
4. XYZ  VWZ
2. Vertical s Thm.
3. Given
4. SAS
Use SAS to explain why
∆ABC  ∆DBC.
Statements
Reasons
1. Given
1. BA  BD
2.
ABC 
DBC
2. Given
3. BC  BC
3. Reflexive Property of
Congruence
4. ABC  DBC
4. SAS
Show that the triangles are congruent for the given value
of the variable.
∆MNO  ∆PQR, when x = 5.
PQ = x + 2
=5+2=7
QR = x = 5
S→ PQ  MN,
S→ QR  NO,
S→ PR  MO
PR = 3x – 9
= 3(5) – 9 = 6
∆MNO  ∆PQR by SSS.
Example: 8
Show that the triangles are congruent for the given value
of the variable.
∆STU  ∆VWX, when y = 4.
ST = 2y + 3
= 2(4) + 3 = 11
TU = y + 3
S→ST  VW
A→T  W
S→TU  WX
=4+3=7
mT = 20y + 12
= 20(4)+12 = 92°
∆STU  ∆VWX by SAS.
Example: 9
Show that ∆ADB  ∆CDB, t = 4.
DA = 3t + 1
= 3(4) + 1 = 13
DC = 4t – 3
= 4(4) – 3 = 13
mD = 2t2
= 2(16)= 32°
DA  DC
Def. of  segments.
ADB  CDB Def. of .
DB  DB
Reflexive Prop. of .
∆ADB  ∆CDB by SAS.
Example: 10
Given: BC ║ AD, BC  AD
Prove: ∆ABD  ∆CDB
Statements
Reasons
1. BC  AD
2. BC || AD
1. Given
2. Given
3. CBD  ABD
3. Alt. Int. s Thm.
4. BD  BD
4. Reflex. Prop. of 
5. ∆ABD  ∆CDB
5. SAS
Example: 11
Given: QP bisects RQS. QR  QS
Prove: ∆RQP  ∆SQP
Statements
Reasons
1. QR  QS
1. Given
2. QP bisects RQS
2. Given
3. RQP  SQP
3. Def. of angle bisector
4. QP  QP
4. Reflexive Prop. of 
5. ∆RQP  ∆SQP
5. SAS