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PLEASE HELP WITH MATH, PROOFS QUESTION, GIVING BRAINLIEST
ANSWER RIGHT PLEASE MY GRADE DEPENDS ON IT

Given: △ABC , DE∥AC
Prove: DABD=ECBE

Drag and drop a statement or reason to each box to complete the proof.

PLEASE HELP WITH MATH PROOFS QUESTION GIVING BRAINLIEST ANSWER RIGHT PLEASE MY GRADE DEPENDS ON IT Given ABC DEAC Prove DABDECBE Drag and drop a statement or re class=
PLEASE HELP WITH MATH PROOFS QUESTION GIVING BRAINLIEST ANSWER RIGHT PLEASE MY GRADE DEPENDS ON IT Given ABC DEAC Prove DABDECBE Drag and drop a statement or re class=

Respuesta :

cdw2014
The Corresponding Angles Postulate states that when parallel lines cross a third line, the angles created are congruent. So the first blank should be BDE=BAC and BED=BCA.

Now we know that the triangles BDE and BAC have two sets of congruent angles. Therefore, the second blank should be AA Similarity Postulate.

Finally, the last blank is Segment Addition Postulate which states that if a point Y is on segment XZ, then XY + YZ = XZ.


ashishdwivedilVT

Using the similarity of triangles [tex]\frac{AD}{BD} =\frac{EC}{EB}[/tex] was proved.

It is given that

In △ABC,

DE∥AC

What is similarity of triangles?

Two triangles are similar if corresponding angles are equal and corresponding sides are in proportion.

Let us consider △BDE and △ABC,

∠B=∠B.....common angles

∠BDE =∠BAC....corresponding angles

so, △BDE∼△ABC

Since △BDE and △ABC are similar so, corresponding sides will be in proportion,

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

Subtract 1 from both sides,

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

[tex]\frac{AD}{BD} =\frac{EC}{EB}[/tex]

Thus, we proved  [tex]\frac{AD}{BD} =\frac{EC}{EB}[/tex] using the similarity of triangles.

To get more about similarity visit:

https://brainly.com/question/2166570

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