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If \(A\) is the midpoint of \(RU\) and \(T\) is the midpoint of \(RN\), prove that \(\triangle RAT \sim \triangle RUN\).
Proof:
| Statements | Reasons | |
| 1 | \(\angle ART = \angle URN\) | |
| 2 | \(RA = AU = \frac{1}{2}RU\) | |
| 3 | \(RT = TN = \frac{1}{2}RN\) | |
| 4 | \(\frac{RA}{RU} = \frac{RT}{RN} = \frac{1}{2}\) | |
| 5 | \(\triangle RAT \sim \triangle RUN\) |
Answer variants:
by ASA (1 and 4)
The sides are proportional from \(2\) and \(3\).
\(T\) is the midpoint of \(RN\)
\(\angle R\) is common in \(\triangle RAT\) and \(\triangle RUN\)
\(A\) is the midpoint of \(RU\)
The sides are proportional from \(1\) and \(2\).
by SAS (1 and 4)
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