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NEED HELP ASAP 70 POINTS

Given: ​​ ​ △ACE,BD¯¯¯¯¯∥AE¯¯¯¯¯ ​

Prove: BA/CB=DE/CD

Drag an expression or phrase to each box to complete the proof.

NEED HELP ASAP 70 POINTS Given: ​​ ​ △ACE,BD¯¯¯¯¯∥AE¯¯¯¯¯ ​ Prove: BA/CB=DE/CD Drag-example-1
NEED HELP ASAP 70 POINTS Given: ​​ ​ △ACE,BD¯¯¯¯¯∥AE¯¯¯¯¯ ​ Prove: BA/CB=DE/CD Drag-example-1
NEED HELP ASAP 70 POINTS Given: ​​ ​ △ACE,BD¯¯¯¯¯∥AE¯¯¯¯¯ ​ Prove: BA/CB=DE/CD Drag-example-2
NEED HELP ASAP 70 POINTS Given: ​​ ​ △ACE,BD¯¯¯¯¯∥AE¯¯¯¯¯ ​ Prove: BA/CB=DE/CD Drag-example-3

2 Answers

4 votes

Ayo, for anyone who needs it -

NEED HELP ASAP 70 POINTS Given: ​​ ​ △ACE,BD¯¯¯¯¯∥AE¯¯¯¯¯ ​ Prove: BA/CB=DE/CD Drag-example-1
User WillZ
by
8.5k points
4 votes

Answer:


Explanation:

Statement Reason

ΔACE, BD║AE Given

∠4≅∠1 and ∠3≅∠2 Corresponding angle postulate

ΔACE is similar to ΔBCD Angle-Angle similarity postulate


(CA)/(CB)=(CE)/(CD) Definition of similar triangles

CA=CB+BA and CE=CD+DE Segment addition postulate


(CB+BA)/(CB)=(CD+DE)/(CD) Substitution property of equality.


(CB)/(CB)+(BA)/(CB)=(CD)/(CD)+(DE)/(CD) Addition of fractions.


1+(BA)/(CB)=1+(DE)/(CD) Simplification of fractions


(BA)/(CB)=(DE)/(CD) Subtraction property of equality.

User Gianna
by
8.3k points
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