Can someone please help me with this geometry problem?

Angle Proof

Given: 1. Ray BF bisects <ABC, m<ABD = m<CBE

Prove: m<DBF = m<EBF

Question

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Respuesta :

tramserran tramserran · Answer 1 · 2018-09-23T02:24:18+02:00

1. ray BF bisects ∠ABC 1. given

m∠ABD = m∠CBE

2. ∠ABF ≅ ∠ CBF 2. angle bisector theorem

3. m∠ABF = m∠CBF 3. definition of congruency

4. m∠ABF = m∠ABD + m∠DBF 4. angle addition postulate

m∠CBF = m∠CBE + m∠EBF

5. m∠ABF = m∠CBE + m∠DBF 5. substitution (s1: m∠ABD = m∠CBE)

6. m∠CBF = m∠CBE + m∠DBF 6. substitution (s3: ∠ABF = ∠ CBF)

7. m∠CBE + m∠DBF = m∠CBE + m∠EBF 7. transitive property

8. m∠DBF = m∠EBF 8. subtraction property

RajarshiG RajarshiG · Answer 2 · 2022-06-23T20:56:07+02:00

From the figure it is clear that both the angles m∠DBF and m∠EBF are equal.

Angle is a geometrical figure formed by two rays meeting at a point.

Given, a ray BF bisects ∠ABC.

Therefore, m∠ABF = m∠CBF

From the figure, it is clear that, m∠ABF = m∠ABD + m∠DBF

Similarly, from the figure, it is clear that, m∠CBF = m∠CBE + m∠EBF

Now, m∠ABF = m∠CBE + m∠DBF ...(1) (Given, m∠ABD = m∠CBE)

Here, m∠CBF = m∠CBE + m∠EBF ...(2) (From the figure)

m∠ABF = m∠CBF

Therefore, right hand side of equation(1) and equation(2) are equal.

Now, m∠CBE + m∠DBF = m∠CBE + m∠EBF

Hence, m∠DBF = m∠EBF

Learn more about an angle here:

https://brainly.com/question/13255546

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