Holt McDougal Geometry 4-Ext Proving Constructions Valid 4-Ext Proving Constructions Valid Holt Geometry Lesson Presentation Lesson Presentation Holt McDougal Geometry

4-Ext Proving Constructions Valid Use congruent triangles to prove constructions valid. Objective

Holt McDougal Geometry 4-Ext Proving Constructions Valid 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.

Holt McDougal Geometry 4-Ext Proving Constructions Valid 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. You have learned that there exists exactly one midpoint on any line segment.

Holt McDougal Geometry 4-Ext Proving Constructions Valid To construct a midpoint, see the construction of a perpendicular bisector on p Remember!

Holt McDougal Geometry 4-Ext Proving Constructions Valid Given: Diagram showing the steps in the construction Prove: CD  AB Example 1: Proving the Construction of a Midpoint

Holt McDougal Geometry 4-Ext Proving Constructions Valid 4. Reflex. Prop. of  5. SSS Steps 2, 3, 4 5. ∆ADC  ∆BDC 6. CPCTC 6. ADC  BDC 7.  s that form a lin. pair are rt. s. 7. ADC and BDC are rt. s 8. Def. of  3. Same compass setting used 2. Same compass setting used Statements 1. Through any two points there is exactly one line. Reasons 8. CD  AB 4. CD  CD 3. AD  BD 2. AC  BC 1. Draw AC, BC. Example 1 Continued

Holt McDougal Geometry 4-Ext Proving Constructions Valid Check It Out! Example 1 Given: Prove: CD is the perpendicular bisector of AB.

Holt McDougal Geometry 4-Ext Proving Constructions Valid Check It Out! Example 1 Continued 4. SSS Steps 2, 3 4. ∆ADC  ∆BDC 5. CPCTC 5. ADC  BDC 6. Reflex. Prop. of  7. SAS Steps 2, 5, 67. ∆ACM and ∆BCM 8. CPCTC 8. AMC  BMC 3. Reflex. Prop. of  2. Same compass setting used Statements 1. Through any two points there is exactly one line. Reasons 6. CM  CM 3. CD  CD 2. AC  BC  AD  BD 1. Draw AC, BC, AD, and BD.

Holt McDougal Geometry 4-Ext Proving Constructions Valid 12. Def. of bisector 11. CPCTC 10. Def. of  9. AMC and BMC are rt. s Statements 9.  s supp.  rt. s Reasons 12. CD bisects AB 11. AM  BM 10. AC  BC Check It Out! Example 1 Continued

Holt McDougal Geometry 4-Ext Proving Constructions Valid Given: diagram showing the steps in the construction Prove: D  A Example 2: Proving the Construction of an Angle

Holt McDougal Geometry 4-Ext Proving Constructions Valid Example 2 Continued Since there is a straight line through any two points, you can draw BC and EF. The same compass setting was used to construct AC, AB, DF, and DE, so AC  AB  DF  DE. The same compass setting was used to construct BC and EF, so BC  EF. Therefore ABC  DEF by SSS, and D  A by CPCTC.

Holt McDougal Geometry 4-Ext Proving Constructions Valid Check It Out! Example 2 Prove the construction for bisecting an angle. Draw BD and CD (through any two points. there is exactly one line). Since the same compass setting was used, AB  AC and BD  CD. AD  AD by the Reflexive Property of Congruence. So ABD  ACD by SSS, and BAD  CAD by CPCTC. Therefore AD bisects BAC by the definition of an angle bisector.

Holt McDougal Geometry 4-Ext Proving Constructions Valid To review the construction of an angle congruent to another angle, see p. 22. Remember!