In the illustration, \(D\) is the midpoint of \( \overline{AC} \), \(E\) is the midpoint of \( \overline{BC} \) and \(F\) is the midpoint of \( \overline{EC} \). Then the pair of similar triangles is:
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\(\triangle DEF \) and \(\triangle ABE \)
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\(\triangle DFC \) and \(\triangle AED \)
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\(\triangle DFC \) and \(\triangle DEF \)
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\(\triangle AED \) and \(\triangle ABE \)
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\(\triangle DEF \) and \(\triangle AED \)
Try to solve it before checking the answer.
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\( \triangle DEF \) and \( \triangle ABE \)
\( D \) is the midpoint of \( \overline{AC} \), then:
\( E \) is the midpoint of \( \overline{BC} \), then:
Luego:
This tells us that the pair of segments \( \overline{AD} \) and \( \overline{DC} \) are proportional to the pair \( \overline{BE} \) and \( \overline{EC} \).
The reciprocal theorem to Thales’ theorem tells us that when this occurs, the segments \(\overline{DE}\) and \( \overline{AB} \) are parallel. Consequently, as angles \( \alpha\) and \( \alpha’ \) are corresponding, they are also congruent.
Similarly, since \( D \) is the midpoint of \( \overline{AC} \), and \(F\) is the midpoint of \( \overline{CE} \), then the segments \( \overline{DF} \) and \( \overline{AE} \) are parallel. Consequently, as the angles \( \beta’ \) and \(\beta’ \) are corresponding, they are congruent.
Finally, by the AA criterion of Triangle Properties of triangles, \( \triangle DEF \) and \( \triangle ABE \) are similar.