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Ch. 6

Proving Constructions Valid. Ch. 6. Lesson Presentation. Holt Geometry. Objective. Use congruent triangles to prove constructions valid. When performing a compass and straight edge construction, the compass setting remains the same width until you change it.

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Ch. 6

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  1. Proving Constructions Valid Ch. 6 Lesson Presentation Holt Geometry

  2. Objective Use congruent triangles to prove constructions valid.

  3. When performing a compass and straight edge construction, the compass setting remains the same width until you change it. This fact allows you to construct a segment congruent to a given segment. You can assume that two distances constructed with the same compass setting are congruent.

  4. The steps in the construction of a figure can be justified by combining: • the assumptions of compass and straightedge constructions, and • the postulates and theorems that are used for proving triangles congruent.

  5. Your figure will be a post-construction drawing, including arcs. You will then have to drawlinesegments connecting points in your figure so that you can create triangles that appear to be congruent. Drawing line segments will be actual steps in your proof. The reasonfor introducing a new line segment is the theorem that states “through any two points there is exactly one line.”

  6. Given: BAC, and AD by construction Prove: AD is the angle bisector of BAC.

  7. 1. 3. AC AB ; CD  BD 2. Draw BDand CD. 4. AD AD 7. AD is the angle bisector of BAC. Example 1 Continued Statements Reasons 1. Given 2. Through any two points there is exactly one line. 3. Same compass setting used 4. Reflex. Prop. of  5. SSS Steps3, 4 5. ∆ADC ∆ADB 6. CPCTE 6. DAC DAB 7.  angles  angle bisector

  8. Check It Out! Example 1 Given: Prove: CDis the perpendicular bisector ofAB.

  9. 2. Draw AC, BC, AD, and BD. 3. AC BC  AD  BD 4. CD CD 7. CM CM Example 1 Continued Statements Reasons 1. Given 1. 2. Through any two points there is exactly one line. 3. Same compass setting used 4. Reflex. Prop. of  5. SSS Steps3, 4 5. ∆ADC ∆BDC 6. CPCTE 6. ACD BCD 7. Reflex. Prop. of  8. ∆ACM ∆BCM 8. SAS Steps 2, 5, 6

  10. Statements Reasons 11. AB DC 13. CDbisects AB 12. AM BM Example 1 Continued 9. AMC BMC 9. CPCTC 10. AMC and BMC are lin. pr. 10. Def. of linear pair 11. 2 ’s in linear pr = ―>  sides 12. CPCTE 13. Def. of bisector 14. CDis the perpendicular bisector ofAB. 14. Def. of  bisector

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