# Example 1A: Compare mBAC and mDAC. Compare the side lengths in ∆ABC and ∆ADC. By the Converse of the Hinge Theorem, mBAC > mDAC. AB = AD AC = AC BC.

## Presentation on theme: "Example 1A: Compare mBAC and mDAC. Compare the side lengths in ∆ABC and ∆ADC. By the Converse of the Hinge Theorem, mBAC > mDAC. AB = AD AC = AC BC."— Presentation transcript:

Example 1A: Compare mBAC and mDAC. Compare the side lengths in ∆ABC and ∆ADC. By the Converse of the Hinge Theorem, mBAC > mDAC. AB = AD AC = AC BC > DC Example 1B: Compare EF and FG. By the Hinge Theorem, EF < GF. Compare the sides and angles in ∆EFH angles in ∆GFH. EH = GH FH = FH mEHF > mGHF mGHF = 180° – 82° = 98°

Example 1C: Find the range of values for k. Step 1 Compare the side lengths in ∆MLN and ∆PLN. By the Converse of the Hinge Theorem, mMLN > mPLN. LN = LN LM = LP MN > PN 5k – 12 < 38 k < 10 Substitute the given values. Step 2 Since PLN is in a triangle, mPLN > 0°. Step 3 Combine the two inequalities. The range of values for k is 2.4 < k < 10. 5k – 12 > 0 k > 2.4 Substitute the given values.

Example 2: Travel Application John and Luke leave school at the same time. John rides his bike 3 blocks west and then 4 blocks north. Luke rides 4 blocks east and then 3 blocks at a bearing of N 10º E. Who is farther from school? Explain. The distances of 3 blocks and 4 blocks are the same in both triangles. The angle formed by John’s route (90º) is smaller than the angle formed by Luke’s route (100º). So Luke is farther from school than John by the Hinge Theorem.

Example 3: Proving Triangle Relationships Write a two-column proof. Given: Prove: AB > CB Proof: StatementsReasons 1. Given 2. Reflex. Prop. of  3. Hinge Thm.

1. Given 2. Def. of Midpoint 3. Def. of  s 4. Conv. of Isoc. ∆ Thm. 5. Hinge Thm. 1. C is the mdpt. of BD m3 > m4, m1 = m2 3. 1  2 5. AB > ED StatementsReasons Example 3B Write a two-column proof. Given: C is the midpoint of BD. Prove: AB > ED m1 = m2 m3 > m4

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