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Given: `DeltaABC` with `bar(DE)` || `bar(AC)` 
Prove: `(AD)/(DB) = (CE)/(EB)`



Statement

Reason

1. `bar(DE)` || `bar(AC)`

given

2. `/_ CAB~=/_EDB` 
    `/_ACB~=/_DEB`

If two parallel lines are cut by a transversal, the corresponding angles are congruent.

3. `DeltaABC ~ DeltaDBE`

AA criterion for similarity

4. 

Corresponding sides of similar triangles are proportional.

5. AB = AD + DB
    CB = CE + EB

segment addition

6. `(AD + DB) /(DB) = (CE + EB) /(EB)`

Substitution Property of Equality

7. `(AD)/(DB) + 1 = (CE)/(EB) + 1`

division

8. `(AD)/(DB) = (CE)/(EB)`

Subtraction Property of Equality

What is the missing statement in the proof? Scroll down to see the entire proof.

Given DeltaABC with barDE barAC Prove ADDB CEEBStatementReason1 barDE barACgiven2 CABEDB ACBDEBIf two parallel lines are cut by a transversal the corresponding class=

Respuesta :

aachen

Given: ΔABC with DE║AC

To Prove: [tex] \frac{AD}{DB} =\frac{CE}{EB} [/tex]

Statements and Reasons

1. DE║AC (given)


2. ∡CAB = ∡EDB and ∡ACB = ∡DEB (If two parallel lines are cut by a transversal, the corresponding angles are congruent.
)

3. ΔABC ≈ ΔDBE (Angle-Angle Similarity theorem)

4. [tex] \frac{AB}{DB} =\frac{CB}{EB} [/tex] (Corresponding sides of similar triangles are proportional.)

5. AB = AD + DB
and CB = CE + EB
(segment addition postulate)



6. [tex] \frac{AD + DB}{DB} =\frac{CE + EB}{EB} [/tex] (Substitution Property of Equality
)

7. [tex] \frac{AD}{DB}\; + 1 =\; \frac{CE}{EB}\; + 1 [/tex]
(Division property)

8. [tex] \frac{AD}{DB} =\frac{CE}{EB} [/tex] (Subtraction Property of Equality
)

So missing statement in the proof was [tex] \frac{AB}{DB} =\frac{CB}{EB} [/tex]

MrRoyal

The missing statement in the proof is [tex]\frac{AB}{DB} = \frac{CB}{EB}[/tex]

How to determine the missing statement

The similarity statement of both triangles is given as:

[tex]\triangle ABC \sim \triangle DBE[/tex]

The above statement means that:

  1. Triangles ABC and DBE are similar
  2. Points A and D are corresponding
  3. Points B and B are corresponding
  4. Points C and E are corresponding

By the corresponding sides of similar triangles., we have the following proportional and equivalent ratio

[tex]AB : DB =CB : EB[/tex]

Express the ratio as fraction

[tex]\frac{AB}{DB} = \frac{CB}{EB}[/tex]

Hence, the missing statement in the proof is [tex]\frac{AB}{DB} = \frac{CB}{EB}[/tex]

Read more about similar triangles at:

https://brainly.com/question/14285697

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