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anonymous
 one year ago
In the figure below, ∠ABC ≅ ∠DEC and ∠GFE ≅ ∠DCE. Point C is the point of intersection between segment AG and segment BD, while point E is the point of intersection between segment AG and segment DF.
anonymous
 one year ago
In the figure below, ∠ABC ≅ ∠DEC and ∠GFE ≅ ∠DCE. Point C is the point of intersection between segment AG and segment BD, while point E is the point of intersection between segment AG and segment DF.

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mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Here is the given figure. dw:1440008861250:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Now let's add the given info.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440008960006:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Do you know how to prove triangles similar using only congruent angles?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0AA triangle postulate?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Correct. Using vertical angles, we can do exactly that.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440010430470:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440010452565:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1By using the theorem: "Vertical angles are congruent." we can show that the vertical angles shown in the figure are congruent. Then we have two angles of one triangle congruent to two angles of the other triangle, and by AA Similarity, we prove the two triangles are similar.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would I put that into a paragraph proof?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1By vertical angles, <DEC =~ to <FEG Since by given <ABC =~ DEC, then by transitive, <ABC =~ <FEG By vertical angles, <DCE =~ <BCA Since by given <GFE =~ <DCE, then by transitive, <BCA =~ <GFE By AA Similarity, ΔABC ~ ΔGEF
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