EDGE Math 8

Inequalities of Two Triangles

Theorem. Converse of the Hinge Theorem (SSS Inequality)
If two sides of a triangle are congruent to the two sides of another triangle and the third side of the first triangle is longer than the third side of the second, then the angle between the pair of congruent sides in the first triangle is greater than the corresponding angle in the second triangle.

Given:	$ΔABC$ and $ΔDEF$ $\bar{AC} ≅ \bar{DF}$ $\bar{BC} ≅ \bar{EF}$ $AB > DE$
Prove:	$m ∠ C > m ∠ F$

Indirect Proof:

Assume

m ∠ C \leq m ∠ F .

Consider two cases.

Case 1.

m ∠ C = m ∠ F,

then

ΔABC ≅ ΔDEF

by the SAS Congruence Postulate.
By CPCTC,

\bar{AB} ≅ \bar{DE};

thus,

AB = DE .

This contradicts the given statement that

AB > DE .

Case 2.

m ∠ C < m ∠ F,

then

AB < DE

by the Hinge Theorem (or SAS Inequality). This contradicts the given statement that

AB > DE .

Therefore, the assumption that $m ∠ C \leq m ∠ F$ must be false, and the conclusion that $m ∠ C > m ∠ F$ must be true.