1. Pearson Unit 1 Topic 4: Congruent Triangles 4-5: Isosceles and Equilateral Triangles Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

2. TEKS Focus:
(5)(C) Use the constructions of congruent segments, congruent angles, angle bisectors, and perpendicular bisectors to make conjectures about geometric relationships.
(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.
(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.
(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.
(6)(D) Verify theorems about the relationships in triangles, including proof of the Pythagorean Theorem, the sum of interior angles, base angles of isosceles triangles, midsegments, and medians, and apply these relationships to solve problems.

3. Vocabulary:
Corollary—a theorem that can be proved easily using another theorem.

4. 1 and 2 are the base angles.
Recall that an isosceles triangle has at least two congruent sides. The congruent sides are called the legs. The vertex angle is the angle formed by the legs. The side opposite the vertex angle is called the base, and the base angles are the two angles that have the base as a side. A
AB and BC are the legs.
AC is the base.
3 is the vertex angle.
1 and 2 are the base angles. C B

5. The Isosceles Triangle Theorem is sometimes stated as “Base angles of an isosceles triangle are congruent.”
Reading Math

6. Example 1: Proving Isosceles Triangle Theorem
CPCTC
CPCTC

7. Example 2: Yes because base angles are congruent
in an isosceles triangle.
??? Yes because corresponding angles R and WVS
are congruent proving RT || WV with transversal RS.
Therefore UV is another transversal proving UVR is
congruent to S. UVR is congruent to R by the
Transitive Property of Congruence. If two base angles
are congruent, then the triangle is isosceles.

8. Example 3: x = 90 – 27 x = 63 Given this right angle. Also 27 since
triangle is isosceles.

9. Example: 4 The mYZX = 180 – 140, so mYZX = 40°.
The length of YX is 20 feet.
Explain why the length of YZ is the same.
The mYZX = 180 – 140, so mYZX = 40°.
Since YZX  X, ∆XYZ is isosceles by the Converse of the Isosceles Triangle Theorem.
Thus YZ = YX = 20 ft.

10. Example: 5 Find mF. mF = mD = x° mF + mD + mA = 180
Isosc. ∆ Thm.
mF + mD + mA = 180
∆ Sum Thm.
Substitute the given values.
x + x + 22 = 180
Simplify and subtract 22
from both sides.
2x = 158
Divide both sides by 2.
x = 79
Thus mF = 79°

11. Example: 6 Find mG. mJ = mG (x + 44) = 3x 44 = 2x x = 22
Isosc. ∆ Thm.
Substitute the given values.
(x + 44) = 3x
Simplify x from both sides.
44 = 2x
Divide both sides by 2.
x = 22
Thus mG = 22° + 44° = 66°.

12. Example: 7 Find mH. mH = mG = x° mH + mG + mF = 180
Isosc. ∆ Thm.
mH + mG + mF = 180
∆ Sum Thm.
Substitute the given values.
x + x + 48 = 180
Simplify and subtract 48
from both sides.
2x = 132
Divide both sides by 2.
x = 66
Thus mH = 66°

13. Example: 8 Find mN and mM. mP = mN (8y – 16) = 6y 2y = 16 y = 8
Isosc. ∆ Thm.
Substitute the given values.
(8y – 16) = 6y
Subtract 6y and add
16 to both sides.
2y = 16
y = 8
Divide both sides by 2.
Thus mN = 6(8) = 48°.
mM + 48° + 48° = 180.
mM = 84°

14. The following corollary and its converse show the connection between equilateral triangles and equiangular triangles.

15. Example: 9 Find the value of x. ∆LKM is equilateral.
Equilateral ∆  equiangular ∆
The measure of each  of
an equiangular ∆ is 60°.
(2x + 32) = 60
2x = 28
Subtract 32 both sides.
x = 14
Divide both sides by 2.

16. Example: 10 Find the value of y. ∆NPO is equiangular.
Equiangular ∆  equilateral ∆
5y – 6 = 4y + 12
Definition of equilateral ∆.
y = 18
Subtract 4y and add 6 to
both sides.

17. Example: 11 Find the value of JL. ∆JKL is equiangular.
Equiangular ∆  equilateral ∆
4t – 8 = 2t + 1
Definition of equilateral ∆.
2t = 9
Subtract 4y and add 6 to both sides.
t = 4.5
Divide both sides by 2.
Thus JL = 2(4.5) + 1 = 10.