SOLVED: 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 segme

Angle bisector theorem proof | Special properties and parts of triangles | Geometry | Khan Academy
Angle bisector theorem proof | Special properties and parts of triangles | Geometry | Khan Academy

Get 5 free video unlocks on our app with code GOMOBILE

Snapsolve any problem by taking a picture.
Try it in the Numerade app?

Solved step-by-step

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:
Triangle ABC in Figure 5.51(a), in which CD bisects angle ACB.
AD and DB are the segments formed by the bisecting ray.
PROVE: AC/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 DC. [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 bisects angle ACB, we have:
EC/AD = CB/DB (by Theorem 5.6.1)
Now, angle L1 = angle L2 because CD bisects angle ACB. Angle L2 = angle L4 (corresponding angles for parallel lines), and angle L3 = angle L4 (alternate interior angles for parallel lines). By the Transitive Property, angle L1 = angle L3. Therefore, triangle LACE is isosceles with EC = AC.
Using substitution, the starred proportion becomes:
Figure 5.51

Solved by verified expert

This problem has been solved!

Try Numerade free for 7 days

02:34

PROVING A THEOREM Write a proof of the In center Theorem (Theorem 6.6$)$ . Given $$\begin{array}{l}{\triangle \mathrm{ABC}, \overline{\mathrm{AD}} \text { bisects } \angle \mathrm{CAB}} \\ {\frac{\mathrm{BD}}{\mathrm{BD}} \text { bisects } \angle \overline{\mathrm{CA}}} \\ {\text { and } \overline{\mathrm{DG}} \perp \overline{\mathrm{CA}}}\end{array}$$ Prove The angle bisectors intersect at $\mathrm{D},$ which is equidistant from $\overline{\mathrm{AB}}, \overline{\mathrm{BC}},$ and $\overline{\mathrm{CA}}$ .

01:30

PROVING A THEOREM Prove that the Circumscribed Angle Theorem (Theorem 10.17) follows from theAngles Outside the Circle Theorem (Theorem 10.16 ).

07:27

Rewrite the paragraph proof of the Hypotenuse-Leg (HL) Congruence Theorem as a two-column proof.Given: $\triangle A B C$ and $\triangle D E F$ are right triangles. $\angle C$ and $\angle F$ are right angles. $\overline{A C} \cong \overline{D F},$ and $\overline{A B} \cong \overline{D E}$Prove: $\triangle A B C \cong \triangle D E F$Proof: On $\triangle D E F$ draw $\overrightarrow{E F}$. Mark $G$ so that $F G=C B .$ Thus $\overline{F G} \cong \overline{C B} .$ From the diagram, $\overline{A C} \cong \overline{D F}$ and $\angle C$ and $\angle F$ are right angles. $\overline{D F} \perp \overline{E G}$ by definition of perpendicular lines. Thus $\angle D F G$ is a right angle, and $\angle D F G \cong \angle C . \triangle A B C \cong \triangle D G F$ by SAS. $\overline{D G} \cong \overline{A B}$ by CPCTC. $\overline{A B} \cong \overline{D E}$ as given. $\overline{D G} \cong \overline{D E}$ by the Transitive Property. By the Isosceles Triangle Theorem $\angle G \cong \angle E . \angle D… 01:38 PROVING A THEOREM Copy and complete the paragraph proof for one part of the Inscribed Quadrilateral Theorem (Theorem 10.13)Given$\odot \mathrm{C}$with inscribed quadrilateral DEFG$\begin{aligned} \text { Prove } & m \angle D=m \angle F=180^{\circ} \\ & m \angle E=m \angle G=180^{\circ} \end{aligned}$By the Arc Addition Postulate (Postulate 10.1$)$,$\mathrm{mEFG}$_______$=360^{\circ}$and$\mathrm{mF} \in \mathrm{D}=\mathrm{mDEF}=360^{\circ}$Using the _________ Theorem,$\mathrm{mEDG}=2 \mathrm{m} \angle \mathrm{F}\mathrm{mEFG}=2 \mathrm{m} \angle \mathrm{D}, \mathrm{mBEF}=2 \mathrm{m} \angle \mathrm{G},$and$\mathrm{m} \mathrm{F} \mathrm{CD}=2 \mathrm{m} \angle \mathrm{E}$. By the Substitution Property of Equality, 2$\mathrm{m} \angle \mathrm{D}$________$=360^{\circ},\$ so_________.Similarly,___________.

Transcript

In the given question, we have been given triangle a b c in which line c d bisects the angle, a c b. We need to prove that a d divided by a c equal to d b, divided by c b. So to solve this question, we first extend the line b c and parallel to c d. So this is the extension helinand. We also extend a line from point a that is parallel to c r, so both the lines meet at point e. Suppose this angle, a a c, is 3, so to prove this equation, we have to use the concept of parallel lines so because, as we know, that c d is parallel to a so. We can write that e c, divided by a d equal to c b. Divided by d v, as we know that angle 1 is congruent to angle 2, because c d, bisect angle, a c band, angle, 1 and 3 is also corresponding…

An exclusive promotional rate that’s yours to keep forever

or

EMAIL