6.5 Inequalities For Two Triangles

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Theorem

If two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle.

Given: Seg. HA is congruent to Seg. KC; Seg. HB is congruent to Seg. KD; m Angle K is greater than m Angle H.

Prove: Seg. CD is greater than Seg. AB

Proof:

Draw Ray KL so that m Angle LKD = m Angle H. On Ray KL take point M so that KM = HA. Then either M is on Seg. CD or M is not on Seg. CD.

In both cases Triangle MKD = Triangle AHB by Hinge Theorem, and MD = AB.

Case 1: M is on Seg. CD

CD is greater than MD (Seg. Add. Post. and Prop. of Ineq.)

CD is greater than AB (Substitution Prop.)

Converse of Hinge Theorem

If two sides of one triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle is larger than the included angle of the second.

Same diagram as shown above

Given: Seg. KC is congruent to Seg. HA; Seg. KD is congruent to HB; CD is greater than AB.

Prove: m Angle K is greater than m Angle H.

Proof:

Assume temporarily that m Angle K is not greater than m Angle H.

Then either m Angle K = m Angle H or m Angle K id less than m Angle H.

Case 1: If m Angle K = m Angle H, then Triangle CKD is congruent to Triangle AHB by the Hinge Theorem. (CD = AB)

Case 2: If m Angle K = m Angle H, then CD is less than AB by the Hinge Theorem.

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