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The proof of Theorem 5.6.3 requires the use of Theorem 5.6.1. THEOREM 5.6.3 ? (The Angle-Bisector Theorem) If a ray bisects one angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the two sides that form the bisected angle. GIVEN: ?ABC in Figure 5.51(a), in which CD bisects ?ACB PROVE: AD/AC = DB/CB PROOF: We begin by extending BC beyond C (there is only one line through B and C) to meet the line drawn through A parallel to CD. [See Figure 5.51(b).] Let E be the point of intersection. (These lines must intersect; otherwise, AE would have two parallels, BC and CD, through point C.) Because CD || EA, we have EC/AD = CB/DB () by Theorem 5.6.1. Now ?1 ? ?2 because CD bisects ?ACB, ?1 ? ?3 (corresponding angles for parallel lines), and ?2 ? ?4 (alternate interior angles for parallel lines). By the Transitive Property, ?3 ? ?4, so ?ACE is isosceles with EC ? AC. Using substitution, the starred () proportion becomes AC/AD = CB/DB or AD/AC = DB/CB (by inversion)

          The proof of Theorem 5.6.3 requires the use of Theorem 5.6.1.

THEOREM 5.6.3 ? (The Angle-Bisector Theorem)
If a ray bisects one angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the two sides that form the bisected angle.

GIVEN: ?ABC in Figure 5.51(a), in which CD bisects ?ACB
PROVE: AD/AC = DB/CB
PROOF: We begin by extending BC beyond C (there is only one line through B and C) to meet the line drawn through A parallel to CD. [See Figure 5.51(b).] Let E be the point of intersection. (These lines must intersect; otherwise, AE would have two parallels, BC and CD, through point C.)
Because CD || EA, we have
EC/AD = CB/DB (*)
by Theorem 5.6.1. Now ?1 ? ?2 because CD bisects ?ACB, ?1 ? ?3 (corresponding angles for parallel lines), and ?2 ? ?4 (alternate interior angles for parallel lines). By the Transitive Property, ?3 ? ?4, so ?ACE is isosceles with EC ? AC. Using substitution, the starred (*) proportion becomes
AC/AD = CB/DB or AD/AC = DB/CB (by inversion)

The proof of Theorem 5.6.3 requires the use of Theorem 5.6.1.

THEOREM 5.6.3 ? (The Angle-Bisector Theorem)
If a ray bisects one angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the two sides that form the bisected angle.

GIVEN: ?ABC in Figure 5.51(a), in which CD bisects ?ACB
PROVE: AD/AC = DB/CB
PROOF: We begin by extending BC beyond C (there is only one line through B and C) to meet the line drawn through A parallel to CD. [See Figure 5.51(b).] Let E be the point of intersection. (These lines must intersect; otherwise, AE would have two parallels, BC and CD, through point C.)
Because CD || EA, we have
EC/AD = CB/DB (*)
by Theorem 5.6.1. Now ?1 ? ?2 because CD bisects ?ACB, ?1 ? ?3 (corresponding angles for parallel lines), and ?2 ? ?4 (alternate interior angles for parallel lines). By the Transitive Property, ?3 ? ?4, so ?ACE is isosceles with EC ? AC. Using substitution, the starred (*) proportion becomes
AC/AD = CB/DB or AD/AC = DB/CB (by inversion)

Added by Andrea R.

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