Page 144 - Math Course 2 (Book 2)

Inequalities: Two Triangles

Proof: In ΔMDL and ΔMDN, LD ≅ ND, MD ≅ MD, and
Statements Reasons ML > MN.

1. X is the midpoint of Given The SSS Inequality allows us to conclude that
MB. m∠LDM > m∠MDN.
2. XM = BX Definition of midpoint

3. △MCX is isosceles Given Answer m∠LDM > m∠MDN
Definition of isosceles
4 CM = CX Write an inequality finding the range of values
triangle containing a using the information in the figure.
5. CB > CM Given
M
6. CB > CX Substitution

7. m∠CXB > m∠CMX

18 1410 (9a + 15)0 16
A. SSS Inequality Theorem D
B. SAS Inequality Theorem
C. Substitution 12 12
D. none of the above
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Answer
By the SSS Inequality, m∠LDM > m∠MDN or
m∠MDN < m∠LDM

MO. 10 - L6b m∠MDN < m∠LDM SSS Inequality

Relationships Between Two 9a + 15 < 141 Substitution
Triangles 9a < 126 Subtract 15 from each side.

a < 14 Divide each side by 9.
Let’s Begin

Also, recall that the measure of any angle is always
greater than 0.
Use SAS Inequality in a Proof
9a + 15 > 0
Example
9a > – 15 Subtract 15 from each
side
Write an inequality relating m∠LDM to m∠MDN 15 5
using the information in the figure. 9a > – or – 3 Divide each side by 9.
9
M
The two inequalities can be written as the
5
compound inequality – < a < 14.
3

18 1410 (9a + 15)0 16
5
D Answer – < a < 14.
3
12 12

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