(26 - 30) Given: DE \approx FG DG \approx EF Prove: \triangle DEF \approx \triangle DGF Statement Reason
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C: In the diagram, let's label point D as the common point for both line segments DE and DG. D: Since DE = FG and DG = EF, we can conclude that DE + DG = FG + EF. E: By the transitive property of equality, we can simplify the equation to DE + DG = EF + FG. Show more…
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