Question

roduction to proof given that (overline{rt} cong overline{wx}), which statement must be true? rt + tw = wx + wx = 2(rx) rt = 2(rx) rt + tw = rx

Explanation:

Step1: Recall congruent segments

If \(\overline{RT} \cong \overline{WX}\), then \(RT = WX\) (by definition of congruent segments).

Step2: Analyze segment addition

Assuming points \(R\), \(T\), \(W\), \(X\) are collinear in order \(R - T - W - X\), by segment addition postulate: \(RT + TW = RW\) and \(WX + TW = RX\)? Wait, no, maybe the correct order is \(R - T - W - X\), so \(RX = RT + TW + WX\)? Wait, no, maybe the first option was supposed to be \(RT + TW = WX + TW\) (since \(RT = WX\), adding \(TW\) to both sides gives \(RT + TW = WX + TW\)). Let's check the options again. The first option (probably a typo, maybe \(RT + TW = WX + TW\)): since \(RT = WX\), adding \(TW\) to both sides (addition property of equality) gives \(RT + TW = WX + TW\). The other options: \(WX = 2(RX)\) or \(RT = 2(RX)\) don't follow from \(RT = WX\) (no info on midpoints). \(RT + TW = RX\) would require \(TW + WX = RX\) (if \(RT = WX\)), but that's only if \(T - W - X\) and \(R - T - W - X\), but the first option (with correction) is valid by adding \(TW\) to congruent segments.

Answer:

The statement that must be true is \(RT + TW = WX + TW\) (assuming the first option's typo is corrected, as it's based on \(RT = WX\) and adding \(TW\) to both sides). If the first option is \(RT + TW = WX + TW\), that's the answer.