Download 4.6 Isosceles Triangles and Right Triangles

4.5 Even Answers
28. Not enough information given.
30. Not enough information given.
4.6 Isosceles Triangles and Right Triangles
Goal 1: How to use properties of isosceles
triangles to solve problems
Goal 2: How to use properties of right triangles
to solve problems in geometry
In lesson 4.1, you learned that if an
isosceles triangle has exactly two
Congruent sides, then these sides
are the legs of the triangle, and the Leg
noncongruent side is the base. The
two angles that have the base as
one side are the base angles.
Leg
Base
angles
Base
4.6 Isosceles Triangles and Right Triangles
Base Angles THM
If two sides of a triangle are congruent ,
then the angles opposite them are congruent.
N
C
Proof : Refer to the figure
H
Given : NC  NY Prove : C  Y
Statements
Y
1. Label H as the
Reasons
1. Ruler Postulate
midpoint of CY .
2. Draw NH
3. CH  HY
4. NH  NH
5. NC  NY
6. NHC  NHY
7. C  Y
2. Two points determine a line.
3. Definition
4. Reflexive Prop. of 
5. Given
6. SSS
7. CPCTC
4.6 Isosceles Triangles and Right Triangles
If two angles of a triangle are congruent,
then the sides opposite them are congruent.
4.6 Isosceles Triangles and Right Triangles
If a triangle is equilateral,
then it is also equiangular.
4.6 Isosceles Triangles and Right Triangles
If a triangle is equiangular,
then it is also equilateral.
34
2
1
An icosahedro n is a 3 - D figure that has 20  ' s
as its faces. Each verte x of each  is shared by 4
other ' s.
An icosahedro n can be used to make a flat
map used to approximat e distances on Earth' s
surface.
What % of Earth' s
surface is occupied
by North America?
34
2
1
An icosahedro n can be used to make a flat
map used to approximat e distances on Earth' s
surface.
4.6 Isosceles Triangles and Right Triangles
HL Congruence THM
If the hypotenuse and leg of one right triangle
are congruent to the hypotenuse and leg of a 2nd
right triangle, then the two triangles are congruent.
The antena is  to the plane containing the
points B, C , D, and O. Each of the stays
A
running from the top of the antenna to
B, C , and D uses the same length of cable.
Is this enough informatio n to conclude
AOB  AOC  AOD ?
D
B
O
C
Yes  By the HL  THM
4.6 Isosceles Triangles and
Right Triangles
HW 4.6/11 - 19odd, 20 - 22, 25 - 33 odd
4.6 Even Answers
20. Since ABD  CBD, AB  BC , and ABE  CBE
by CPCTC ; BE  BE by the Reflexive Prop.
ABE  CBE by SAS , and BAE  BCE
by CPCTC.
22. Since ABD  CBD  CDG  CFG and all are
equilateral; so ABD  DBC  DGC  FGC
by CPCTC and because an equilateral  is equiangular.
Then mABD  mDBC  mDGC  mFGC by
+ Prop. of =. Then mABC  mDGF by the
 + Post. and ABC  DGF by definition.