Exercise 47 Page 342

Let's consider the theorem we want to prove, which is the Triangle Inequality Theorem.

Triangle Inequality Theorem

The sum of the lengths of any two sides of a triangle is greater than the length of the third side. lAB+BC > AC AC+BC > AB AB+AC > BC

Let's prove this theorem using an indirect proof! We will begin by drawing an arbitrary triangle with vertices A, B, and C.

We will prove the first expression. Since we are doing an indirect proof, we will temporarily assume that AB+BC is less than or equal to AC. Temporary Assumption AB+BC≤ AC Now, we will extend BC to a point D such that we create an isosceles triangle with vertices A, B, and D, and where AB ≅ DB.

Since AB and DB are congruent, by the definition of congruent segments we know that their measures are equal. Knowing this, we can substitute AB for DB in the inequality and use the Segment Addition Postulate to write DB+BC as DC.

AB+BC≤ AC

Substitute

AB= DB

DB+BC≤ AC

Substitute

DB+BC= DC

DC≤ AC

We obtained that DC is less than or equal to AC. If DC=AC, then △ADC is an isosceles triangle. By the Isosceles Triangle Theorem, if two sides of a triangle are congruent, then the angles opposite them are congruent. By the definition of congruent angles, this means that ∠DAC and ∠ADC have the same measure. m∠ DAC = m∠ ADC By the Triangle Longer Side Theorem, if one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side. Consequently, if DC< AC, then m∠DAC < m∠ADC.

Therefore, if DC ≤ AC, we have that m∠ DAC ≤ m∠ ADC. Let's now use the Angle Addition Postulate to rewrite m∠DAC as the sum of m∠DAB and m∠BAC.

This means that we can substitute m∠ DAB + m∠ BAC for m∠ DAC in our previous inequality. m∠ DAC ≤ m∠ ADC ⇓ m∠DAB + m∠BAC ≤ m∠ADC Since △ABD is isosceles, we know that m∠DAB=m∠ADB. m∠DAB +m& ∠BAC ≤ m∠ADC &⇓ m∠ADB +m& ∠BAC ≤ m∠ADC We can also notice that ∠ADB and ∠ADC are the same angle. Therefore, by the Reflexive Property of Congruence, they have the same measure. m∠ADB +m& ∠BAC ≤ m∠ADC &⇓ m∠ADC +m& ∠BAC ≤ m∠ADC Let's finally subtract m∠ ADC from both the left-hand side and the right-hand side of the obtained inequality.

m∠ADC + m∠BAC ≤ m∠ADC

SubIneq

LHS-m∠ADC≤RHS-m∠ADC

m∠BAC ≤ 0 *

Since ∠ BAC is an interior angle of a triangle, it cannot have a negative measure. Therefore, we have reached a contradiction! This contradiction came from assuming that AB+BC is less than or equal to AC. Therefore, AB+BC must be greater than AC. AB+BC> AC ✓ By following a similar procedure, we can prove the other two inequalities of the Triangle Inequality Theorem.

Suppose	Extend	Contradiction	Conclusion
AB+BC ≤ AC	BC to DB such that DB≅ AB	m∠BAC ≤ 0 *	AB+BC > AC ✓
AC+BC ≤ AB	AC to AD such that CD≅ BC	m∠ABC ≤ 0 *	AC+BC > AB ✓
AB+AC ≤ BC	AB to BD such that AD≅ AC	m∠ACB ≤ 0 *	AB+AC > BC ✓