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# bisector-and-perpendicular-line-lesson-19.pptx

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# bisector-and-perpendicular-line-lesson-19.pptx

Bisector and Perpendicular line

Bisector and Perpendicular line

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### bisector-and-perpendicular-line-lesson-19.pptx

1. 1. ๐ซ๐๐๐๐๐๐๐๐๐ ๐๐ ๐ซ๐๐๐๐๐๐๐ ๐๐๐๐ ๐ ๐ท๐๐๐๐ ๐๐ ๐ ๐ณ๐๐๐ The distance from a point to a line is the length of the perpendicular segment from the point to the line. If the point lies on the line, the distance is zero. P C A B The distance from point P to AB is PC. Here PC โฅ AB
2. 2. Theorem 35 A point on the bisector of an angle is equidistant from the sides of the angle. A D G F E C B Given: ๐ฉ๐ซ bisects โ ๐จ๐ฉ๐ช, G lies on ๐ฉ๐ซ Prove: ๐ฎ๐ฌ = ๐ฎ๐ญ
3. 3. A D G F E C B Given: ๐ฉ๐ซ bisects โ ๐จ๐ฉ๐ช, G lies on ๐ฉ๐ซ Prove: ๐ฎ๐ฌ = ๐ฎ๐ญ ๐๐ญ๐๐ญ๐๐ฆ๐๐ง๐ญ๐ฌ ๐๐๐๐ฌ๐จ๐ง๐ฌ 1. ๐๐ bisects โ ๐๐๐, G lies on ๐๐ ๐๐ข๐ฏ๐๐ง 2. โ ๐๐๐ โ โ ๐๐๐ ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐๐ง๐ ๐ฅ๐ ๐๐ข๐ฌ๐๐๐ญ๐จ๐ซ 3. ๐๐ โฅ ๐๐ ๐๐ โฅ ๐๐ ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐ญ๐ก๐ ๐๐ข๐ฌ๐ญ๐๐ง๐๐ ๐๐ซ๐จ๐ฆ ๐ ๐๐จ๐ข๐ง๐ญ ๐ญ๐จ ๐ ๐๐ข๐ง๐ 4. โ ๐๐๐ ๐๐ง๐ โ ๐๐๐ ๐๐ซ๐ ๐ซ๐ข๐ ๐ก๐ญ ๐๐ง๐ ๐ฅ๐๐ฌ. ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐๐๐ซ๐ฉ๐๐ง๐๐ข๐๐ฎ๐ฅ๐๐ซ 5. ๐๐ โ ๐๐ ๐๐๐๐ฅ๐๐ฑ๐ข๐ฏ๐ ๐๐ซ๐จ๐ฉ๐๐ซ๐ญ๐ฒ ๐จ๐ ๐๐จ๐ง๐ ๐ซ๐ฎ๐๐ง๐๐ 6. โ๐๐๐ ๐๐ง๐ โ๐๐๐ ๐๐ซ๐ ๐ซ๐ข๐ ๐ก๐ญ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐๐ฌ. ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐๐ข๐ ๐ก๐ญ ๐๐ซ๐ข๐๐ง๐ ๐ฅ๐๐ฌ 7. โ๐๐๐ โ โ๐๐๐ ๐๐ฒ๐ฉ๐จ๐ญ๐๐ง๐ฎ๐ฌ๐ โ ๐๐๐ฎ๐ญ๐ ๐๐ง๐ ๐ฅ๐ 8. ๐๐ โ ๐๐ ๐๐๐๐๐ 9. ๐๐ = ๐๐ ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐๐จ๐ง๐ ๐ซ๐ฎ๐๐ง๐ญ ๐๐๐ ๐ฆ๐๐ง๐ญ๐ฌ proof:
4. 4. Theorem 36 A point equidistant from the sides of an angle lies on the bisector of an angle A D G F E C B Given: ๐บ๐ธ โฅ ๐ต๐ด; ๐บ๐น โฅ ๐ต๐ถ, ๐บ๐ธ = ๐บ๐น Prove: ๐บ lies on the bisector of โ ๐ด๐ต๐ถ. ๐ท๐๐๐๐๐๐๐๐๐๐๐๐ ๐ฉ๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐บ๐๐๐๐๐๐ Recall that the perpendicular bisector of a segment is a line, ray, segment, or plane that is perpendicular to the segment at its midpoint.
5. 5. ๐ป๐๐๐๐๐๐ ๐๐ A point on the perpendicular bisector of a segment is equidistant from the endpoint of a segment. C P D B A Given: ๐ช๐ซ โฅ ๐๐๐๐๐๐๐๐ ๐๐ ๐จ๐ฉ, ๐ท lies on ๐ช๐ซ. Prove: ๐ท๐จ = ๐ท๐ฉ
6. 6. C P D B A Given: ๐ช๐ซ โฅ ๐๐๐๐๐๐๐๐ ๐๐ ๐จ๐ฉ, ๐ท lies on ๐ช๐ซ. Prove: ๐ท๐จ = ๐ท๐ฉ ๐๐๐๐๐: ๐๐ก๐๐ก๐๐๐๐๐ก๐  ๐๐๐๐ ๐๐๐  1. ๐ถ๐ท โฅ bisector of ๐ด๐ต, ๐ lies on ๐ถ๐ท. ๐บ๐๐ฃ๐๐ 2. ๐ท๐๐๐ค ๐ด๐ ๐๐๐ ๐ต๐. ๐ฟ๐๐๐ ๐๐๐ ๐ก๐ข๐๐๐ก๐ 3. ๐ถ๐ท โฅ ๐ด๐ต; ๐ท ๐๐  ๐กโ๐ ๐๐๐๐๐๐๐๐ก ๐๐ ๐ด๐ต. ๐ท๐๐๐๐๐๐ก๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐๐ข๐๐๐ ๐ต๐๐ ๐๐๐ก๐๐ 4. โ ๐๐ท๐ด ๐๐๐ โ ๐๐ท๐ต ๐๐๐ ๐๐๐โ๐ก ๐๐๐๐๐. ๐ท๐๐๐๐๐๐ก๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐๐ข๐๐๐ 5. โ๐๐ท๐ด ๐๐๐ โ๐๐ท๐ต ๐๐๐ ๐๐๐โ๐ก ๐๐๐๐๐. ๐ท๐๐๐๐๐๐ก๐๐๐ ๐๐ ๐๐๐โ๐ก ๐๐๐๐๐๐๐๐๐  6. ๐ด๐ท โ ๐ต๐ท ๐ท๐๐๐๐๐๐ก๐๐๐ ๐๐ ๐ ๐๐๐๐๐๐๐๐ก 7. ๐๐ท โ ๐๐ท ๐๐๐๐๐๐ฅ๐๐ฃ๐ ๐๐๐๐๐๐๐ก๐ฆ ๐๐ ๐ถ๐๐๐๐๐ข๐๐๐๐ 8. โ๐๐ท๐ด โ โ๐๐ท๐ต ๐ฟ๐๐ โ ๐ฟ๐๐ ๐โ๐๐๐๐๐ 9. ๐๐ด โ ๐๐ต ๐ถ๐๐ถ๐๐ถ 10. ๐๐ด = ๐๐ต ๐ท๐๐๐๐๐๐ก๐๐๐ ๐๐ ๐ถ๐๐๐๐๐ข๐๐๐ก ๐๐๐๐๐๐๐ก๐
7. 7. ๐ป๐๐๐๐๐๐ ๐๐ A point equidistant from the endpoint of a segment lies on the perpendicular bisector of a segment. C P D B A Given: ๐๐ = ๐๐ Prove: P lies on the perpendicular bisector of AB.
8. 8. C P D B A Given: ๐๐ = ๐๐ Prove: P lies on the perpendicular bisector of AB. ๐๐๐๐๐: ๐บ๐๐๐๐๐๐๐๐๐ ๐น๐๐๐๐๐๐ 1. ๐ณ๐๐ ๐ซ ๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐จ๐ฉ ๐ป๐๐๐๐๐๐ ๐. ๐ฌ๐๐๐๐ ๐๐๐๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐ 2. ๐จ๐ซ โ ๐ฉ๐ซ ๐ซ๐๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐ด๐๐๐๐๐๐๐ 3. ๐ซ๐๐๐ ๐๐๐๐ ๐ท๐ซ ๐ณ๐๐๐ ๐ท๐๐๐๐๐๐๐๐ 4. ๐ท๐ซ โ ๐ท๐ซ ๐น๐๐๐๐๐๐๐๐ ๐ท๐๐๐๐๐๐๐ ๐๐ ๐ช๐๐๐๐๐๐๐๐๐ 5. ๐ท๐จ = ๐ท๐ฉ ๐ฎ๐๐๐๐ 6. โ๐ท๐ซ๐จ โ โ๐ท๐ซ๐ฉ ๐บ๐บ๐บ ๐ท๐๐๐๐๐๐๐๐ 7. โ ๐ท๐ซ๐จ โ โ ๐ท๐ซ๐ฉ ๐ช๐ท๐ช๐ป๐ช 8. โ ๐ท๐ซ๐จ ๐๐๐ โ ๐ท๐ซ๐ฉ ๐๐๐๐ ๐ ๐๐๐๐๐๐ ๐๐๐๐. ๐ซ๐๐๐๐๐๐๐๐๐ ๐๐ ๐ณ๐๐๐๐๐ ๐ท๐๐๐ 9.โ ๐ท๐ซ๐จ ๐๐๐ โ ๐ท๐ซ๐ฉ ๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐๐. ๐ณ๐๐๐๐๐ ๐ท๐๐๐ ๐ท๐๐๐๐๐๐๐๐ 10. โ ๐ท๐ซ๐จ ๐๐๐ โ ๐ท๐ซ๐ฉ ๐๐๐ ๐๐๐๐๐ ๐๐๐๐๐ ๐ป๐๐๐๐๐๐ ๐๐. ๐ฐ๐ ๐๐๐ ๐๐๐๐๐๐ ๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐๐, ๐๐๐๐ ๐๐๐๐ ๐๐ ๐ ๐๐๐๐๐ ๐๐๐๐๐.
9. 9. ๐ป๐๐๐๐๐๐ ๐๐ If a line contains two points each of which is equidistant from the endpoints of the segment, then the line is the perpendicular bisector of the segment. C P D B A Given: AP=BP, AC=BC Prove: PD is the โฅ bisector of AB.
10. 10. C P D B A Given: AP=BP, AC=BC Prove: PD is the โฅ bisector of AB. ๐๐ญ๐๐ญ๐๐ฆ๐๐ง๐ญ๐ฌ ๐๐๐๐ฌ๐จ๐ง๐ฌ 1. ๐๐ โ ๐๐, ๐๐ โ ๐๐ ๐๐ข๐ฏ๐๐ง ๐๐ง๐ ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐๐จ๐ง๐ ๐ซ๐ฎ๐๐ง๐ญ ๐๐๐ ๐ฆ๐๐ง๐ญ๐ฌ 2. ๐๐ โ ๐๐ ๐๐๐๐ฅ๐๐ฑ๐ข๐ฏ๐ ๐๐ซ๐จ๐ฉ๐๐ซ๐ญ๐ฒ ๐จ๐ ๐๐จ๐ง๐ ๐ซ๐ฎ๐๐ง๐๐ 3. โ๐๐๐ โ โ๐๐๐ ๐๐๐ ๐๐จ๐ฌ๐ญ๐ฎ๐ฅ๐๐ญ๐ 4. โ ๐๐๐ โ โ ๐๐๐ ๐๐๐๐๐ 5. ๐๐ โ ๐๐ ๐๐๐๐ฅ๐๐ฑ๐ข๐ฏ๐ ๐๐ซ๐จ๐ฉ๐๐ซ๐ญ๐ฒ ๐จ๐ ๐๐จ๐ง๐ ๐ซ๐ฎ๐๐ง๐๐ 6. โ๐๐๐ โ โ๐๐๐ ๐๐๐ ๐๐จ๐ฌ๐ญ๐ฎ๐ฅ๐๐ญ๐ 7. ๐๐ โ ๐๐ ๐๐๐๐๐ 8. ๐ ๐ข๐ฌ ๐ญ๐ก๐ ๐ฆ๐ข๐๐ฉ๐จ๐ข๐ง๐ญ ๐จ๐ ๐๐ ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐๐ข๐๐ฉ๐จ๐ข๐ง๐ญ 9. โ ๐๐๐ โ โ ๐๐๐ ๐๐๐๐๐ 10.โ ๐๐๐ ๐๐ง๐ โ ๐๐๐ ๐๐จ๐ซ๐ฆ ๐ ๐ฅ๐ข๐ง๐๐๐ซ ๐๐ข๐ซ๐ฆ ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐๐ข๐ง๐๐๐ซ ๐๐๐ข๐ซ 11. โ ๐๐๐ ๐๐ง๐ โ ๐๐๐ ๐๐ซ๐ ๐ฌ๐ฎ๐ฉ๐ฉ๐ฅ๐๐ฆ๐๐ง๐ญ๐๐ซ๐ฒ ๐๐ข๐ง๐๐๐ซ ๐๐๐ข๐ซ ๐๐จ๐ฌ๐ญ๐ฎ๐ฅ๐๐ญ๐ 12. โ ๐๐๐ ๐๐ง๐ โ ๐๐๐ ๐๐ซ๐ ๐ซ๐ข๐ ๐ก๐ญ ๐๐ง๐ ๐ฅ๐๐ฌ ๐๐ก๐๐จ๐ซ๐๐ฆ ๐๐. ๐๐ ๐ญ๐ฐ๐จ ๐๐ง๐ ๐ฅ๐๐ฌ ๐๐ซ๐ ๐๐จ๐ญ๐ก ๐๐จ๐ง๐ ๐ซ๐ฎ๐๐ง๐ญ ๐๐ง๐ ๐ฌ๐ฎ๐ฉ๐ฉ๐ฅ๐๐ฆ๐๐ง๐ญ๐๐ซ๐ฒ, ๐ญ๐ก๐๐ง ๐๐๐๐ก ๐ข๐ฌ ๐ ๐ซ๐ข๐ ๐ก๐ญ ๐๐ง๐ ๐ฅ๐ 13. ๐๐ ๐ข๐ฌ ๐ญ๐ก๐ โฅ ๐๐ข๐ฌ๐๐๐ญ๐จ๐ซ ๐จ๐ ๐๐ ๐๐๐๐ข๐ง๐ข๐ญ๐ข๐จ๐ง ๐จ๐ ๐ฉ๐๐ซ๐ฉ๐๐ง๐๐ข๐๐ฎ๐ฅ๐๐ซ ๐๐ข๐ฌ๐๐๐ญ๐จ๐ซ proof:
11. 11. ๐ป๐๐๐๐๐๐ ๐๐ In a plane, through a given points on a line, there is exactly one line perpendicular to the line. R S M P Q Given: MQ and point P on MQ Prove: a. There is at least on line PR โฅ MQ b. There is only one line PR โฅ MQ Part 1: Show that there is one line ๐ท๐น โฅ ๐ด๐ธ through point P. Through the end point of ๐ท๐ธ , ray ๐ท๐น can be constructed in such a way that ๐โ ๐น๐ท๐ธ = ๐๐. โ ๐น๐ท๐ธis therefore a right angle and hence, by the definition of perpendicular, ๐ท๐น โฅ ๐ด๐ธ . Since rays ๐ท๐น โฅ ๐ท๐ธ are contained in ๐ท๐น ๐๐๐ ๐ด๐ธ , respectively, therefore ๐ท๐น โฅ ๐ด๐ธ .
12. 12. Part 2: Show that ๐ท๐น ๐๐ ๐๐๐ ๐๐๐๐ ๐๐๐๐ โฅ ๐ด๐ธ through P. There is only one line ๐ท๐น โฅ ๐ด๐ธ or other than ๐ท๐น there is another line ๐ท๐น โฅ ๐ด๐ธ . Assume that there is another line ๐ท๐บ โฅ ๐ด๐ธ . Then by the definition of perpendicular, โ ๐บ๐ท๐ธ is a right angle and therefore ๐โ ๐น๐ท๐ธ = ๐๐. This contradicts the Angle Construction Postulate. Since the assumption leads to contradiction of postulate, it must be false. Therefore, there is only one line ๐ท๐น โฅ ๐ด๐ธ . R S M P Q Given: MQ and point P on MQ Prove: a. There is at least on line PR โฅ MQ b. There is only one line PR โฅ MQ
13. 13. ๐ป๐๐๐๐๐๐ ๐๐ There is one and only one line perpendicular to a given line through an external point. For Example: ๐ผ๐ ๐กโ๐ ๐๐๐๐ข๐๐, ๐๐๐๐๐ก ๐ ๐๐๐๐  ๐๐ ๐ป๐ท, ๐กโ๐ ๐๐๐ ๐๐๐ก๐๐ ๐๐ โ ๐พ๐ป๐. If ๐พ๐ป = 4๐ฅ โ 15 and ๐๐ป = 2๐ฅ + 3. Find ๐๐ป. SOLUTION: ๐ฒ๐ฏ = ๐ต๐ฏ 4๐ฅ โ 15 = 2๐ฅ + 3 Substitute the value of ๐พ๐ป and ๐๐ป. 4๐ฅ โ 2๐ฅ = 3 + 15 Combine similar terms. 2๐ฅ = 18 Simplify. 2๐ฅ 2 = 18 2 Divide both side by 2. ๐ฅ = 9 ๐๐ป = 2๐ฅ + 3 ๐๐ป = 2 9 + 3 Substitute the value of x ๐๐ป = 18 + 3 Simplify. ๐๐ป = 21 H D N M K
14. 14. EXAMPLE: ๐ผ๐ ๐กโ๐ ๐๐๐๐ข๐๐, ๐ท๐ ๐๐  ๐กโ๐ ๐๐๐๐๐ ๐๐๐ ๐๐๐ก๐๐ ๐๐ โ ๐๐ท๐. ๐๐ฟ = 5๐ฅ โ 12, ๐๐ฟ = 2๐ฅ + 3, ๐ท๐ = 2๐ฆ โ 3, ๐๐๐ ๐ท๐ = ๐ฆ + 12. Find the perimeter of ๐๐ท๐๐ฟ. ๐๐ฟ = ๐๐ฟ ๐ท๐ = ๐ท๐ 5๐ฅ โ 12 = 2๐ฅ + 3 2๐ฆ โ 3 = ๐ฆ + 12 5๐ฅ โ 2๐ฅ = 3 + 12 2๐ฆ โ ๐ฆ = 12 + 3 3๐ฅ = 15 ๐ฆ = 15 ๐ฅ = 5 ๐๐ฟ = 5๐ฅ โ 12 ๐๐ฟ = 2๐ฅ + 3 ๐ท๐ = 2๐ฆ โ 3 ๐ท๐ = ๐ฆ + 12 ๐๐ฟ = 5 5 โ 12 ๐๐ฟ = 2(5) + 3 ๐ท๐ = 2(15) โ 3 ๐ท๐ = 15 + 12 ๐๐ฟ = 25 โ 12 ๐๐ฟ = 10 + 3 ๐ท๐ = 30 โ 3 ๐ท๐ = 27 ๐๐ฟ = 13 ๐๐ฟ = 13 ๐ท๐ = 27 Since, we are looking for the perimeter of ๐๐ท๐๐ฟ. Just add the length of the sides that we get. ๐ = ๐๐ฟ + ๐๐ฟ + ๐ท๐ + ๐ท๐ ๐ = 13 + 13 + 27 + 27 ๐ = 80 T L D Q M