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General Information

Related Benchmark: MAFS.912.G-SRT.2.5

Reporting Category: Geometry: Similarity, Right Triangles, & Trigonometry

Type: ETC: Editing Task Choice

Difficulty: N/A

Question:

There are three highlights in the paragraph to show blanks in the proof. For each highlight, click on the word or phrase to fill in the blank.

Possible Answer:

It is given that $begin mathsize 12px style stack A D with bar on top end style$ bisects $begin mathsize 12px style angle E A C end style$ , and $begin mathsize 12px style angle F D A approximately equal to angle B D A end style$ . Since $begin mathsize 12px style stack A D with bar on top end style$ bisects $begin mathsize 12px style angle E A C end style$ , then $begin mathsize 12px style angle D A E approximately equal to angle D A C end style$ from the definition of angle bisector. $begin mathsize 12px style stack A D with bar on top approximately equal to stack A D with bar on top end style$ by the reflexive property. $begin mathsize 12px style increment bold italic D bold italic A bold italic F end style$ $begin mathsize 12px style approximately equal to increment bold italic D bold italic A bold italic B end style$ because of ASA. Therefore, $begin mathsize 12px style stack F D with bar on top approximately equal to stack B D with bar on top end style$ because corresponding parts of congruent triangles are congruent.