1. Apply properties of isosceles and equilateral triangles.
Objectives
Apply properties of isosceles and equilateral triangles.
Then/Now

2. legs of an isosceles triangle vertex angle base angles
Vocabulary

3. Recall that an isosceles triangle has at least two congruent sides
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.
3 is the vertex angle.
1 and 2 are the base angles.

4. Concept

6. I DO: Finding the Measure of an Angle
Find mG.
mJ = mG
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°.

7. I DO: Finding the Measure of an Angle
Find mF.
mF = mD = x°
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°

8. You Do! Example 2A
Find mH.
mH = mG = x°
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°

9. You Do! Example 2B
Find mN.
mP = mN
Isosc. ∆ Thm.
Substitute the given values.
(8y – 16) = 6y
Subtract 6y and add 16 to both sides.
2y = 16
Divide both sides by 2.
y = 8
Thus mN = 6(8) = 48°.

10. I Do: Using Properties of Equilateral Triangles
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.

11. I DO: Using Properties of Equilateral Triangles
Find the value of y.
∆NPO is equiangular.
Equiangular ∆  equilateral ∆
Definition of equilateral ∆.
5y – 6 = 4y + 12
Subtract 4y and add 6 to both sides.
y = 18

12. You Do! Example 3
Find the value of JL.
∆JKL is equiangular.
Equiangular ∆  equilateral ∆
Definition of equilateral ∆.
4t – 8 = 2t + 1
Subtract 4y and add 6 to both sides.
2t = 9
t = 4.5
Divide both sides by 2.
Thus JL = 2(4.5) + 1 = 10.

13. Subtract 60 from each side. Answer: mR = 60 Divide each side by 2.
Find Missing Measures
A. Find mR.
Since QP = QR, QP  QR. By the Isosceles Triangle Theorem, base angles P and R are congruent, so mP = mR . Use the Triangle Sum Theorem to write and solve an equation to find mR.
Triangle Sum Theorem
mQ = 60, mP = mR
Simplify.
Subtract 60 from each side.
Answer: mR = 60
Divide each side by 2.
Example 2

14. A. Find mT.
A. 30°
B. 45°
C. 60°
D. 65°
Example 2a

15. B. Find TS.
A. 1.5
B. 3.5
C. 4
D. 7
Example 2b

16. Find the value of each variable.
A. x = 20, y = 8
B. x = 20, y = 7
C. x = 30, y = 8
D. x = 30, y = 7
Example 3

17. Exit Slip (Pick one)
Find each angle measure.
1. mR and mP
Find each value.
3. x y
5. x
28°
124° 6 20
26°