The following steps comprise a proof that the inscribed circle of a triangle alw
ID: 3008053 • Letter: T
Question
The following steps comprise a proof that the inscribed circle of a triangle always exists. Fill in the reasons for each of the steps indicated, Given triangle Delta ABC. 1. Let AX' be the angle bisector of A. 2. AX'meets BC at a point X. 3. Let BY' be the angle bisector of angle B. 4. BY' meets AX at a point O. 5. O is in the interior of the triangle. (Def of interior of triangle, def, of interior of angle, prop. 3.8, and del, of angle bisector.) 6. Drop perpendiculars from O to lines AB, AC, and BC, with feet P, Q, and R, respectively. 7. P is on AB, Q is on AC, and R is on BC. 8. PAO QAO, PBO RBO. 9. OA OA and OB OB. 10. OPA OQA and OPB ORB. 11. Delta OPA Delta OQA and Delta OPB Delta ORB. 12. OQ OP OR. (Corresponding sides and Axiom C-2.) 13. The circle gamma centered at O with radius OP contains points P, Q, and R, (def of circle) 14. let Z be any point on the triangle distinct from P, Q, and R, say, WLOG, Z is on AB, (def of triangle. Axiom B-2). 15. OZ > OP. 16. P, Q, and R are the only points common to both y and the triangle; gamma is called the inscribed circle. (def of circle) 17. OC OC 18. Delta OCQ Delta OCR 19. OCB OCA, so OC bisects angle C.Explanation / Answer
There the reasons for each step will be as follows :
Step REason
1. Given
2. Given
3. Given
4. Given
6. by construction
7. By construction
8. Definition of angle bisectors
9. Reflexive property
10. Definition of perpendiculars
11. Using AAS (Angle-Angle-Side) congruency
15. Perpendicular is the shortest distance
17. Reflexive property
18. By SSS congruency
19. By CPCTP (Corresponding parts of corresponding triangles)
Answer
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