Question

$\overline{vw} \parallel \overline{xz}$. find $wy$.

$wy = \square$

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Explanation:

Step1: Identify Similar Triangles

Since \(\overline{VW} \parallel \overline{XZ}\), by the Basic Proportionality Theorem (Thales' theorem), \(\triangle VWY \sim \triangle XZY\) (similar triangles) because corresponding angles are equal (alternate interior angles and common angle at \(Y\)).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. So, \(\frac{VZ}{ZY}=\frac{WY}{XY + WX}\)? Wait, no. Wait, \(VZ = 46\), \(ZY = 23\), \(WX = 28\), let \(XY = x\), \(WY = 28 + x\). Wait, actually, the sides: \(VZ\) and \(ZY\) are on the base, \(WY\) and \(XY\) (wait, no, \(WY\) is the side, and \(XZ\) is parallel to \(VW\), so the ratio of \(VZ\) to \(ZY\) should equal the ratio of \(WY\) to \(XY\)? Wait, no, let's label the sides correctly.

Wait, \(\triangle VWY\) and \(\triangle XZY\): \(VZ = 46\), \(ZY = 23\), so \(VZ/ZY = 46/23 = 2\). Then, the ratio of \(WY\) to \(XY\) should be 2? Wait, no, \(WY\) is the side from \(W\) to \(Y\), and \(XY\) is from \(X\) to \(Y\), and \(WX\) is from \(W\) to \(X\). Wait, actually, \(WY = WX + XY\), and \(WX = 28\), let \(XY = x\), so \(WY = 28 + x\). Then, since \(\triangle VWY \sim \triangle XZY\), the ratio of corresponding sides: \(\frac{VZ}{ZY}=\frac{WY}{XY}\). Wait, \(VZ = 46\), \(ZY = 23\), so \(46/23 = 2\). So \(2=\frac{WY}{XY}\), but \(WY = WX + XY = 28 + XY\). Wait, no, maybe I mixed up the triangles.

Wait, actually, the correct proportion is \(\frac{VZ}{ZY}=\frac{WY}{XY}\)? No, wait, \(\triangle VZY\) and \(\triangle XZY\)? No, \(\overline{VW} \parallel \overline{XZ}\), so angle at \(Z\): \(\angle VZW\) and \(\angle XZY\) are alternate interior angles? Wait, maybe better to use the ratio of the segments on the base: \(VZ = 46\), \(ZY = 23\), so the ratio of \(VZ\) to \(ZY\) is \(46/23 = 2\). Therefore, the ratio of \(WY\) to \(XY\) should be 2? Wait, no, \(WY\) is the side, and \(XY\) is part of it, with \(WX = 28\). Wait, let's let \(XY = x\), then \(WY = 28 + x\). Then, since \(\triangle VWY \sim \triangle XZY\), \(\frac{VZ}{ZY}=\frac{WY}{XY}\). So \(\frac{46}{23}=\frac{28 + x}{x}\). Wait, \(46/23 = 2\), so \(2=\frac{28 + x}{x}\). Solving: \(2x = 28 + x\) → \(x = 28\). Then \(WY = 28 + 28 = 56\)? Wait, that makes sense. Wait, because \(VZ\) is twice \(ZY\) (46 is twice 23), so \(WY\) should be twice \(XY\). But \(WX\) is 28, so \(XY\) is also 28, so \(WY = 28 + 28 = 56\).

Wait, let's check again. The triangles are similar, so the ratio of the bases \(VZ/ZY = 46/23 = 2\), so the ratio of the heights (or the sides) should also be 2. So \(WY/XY = 2\). But \(WY = WX + XY = 28 + XY\), so \(28 + XY = 2 \times XY\) → \(XY = 28\), so \(WY = 28 + 28 = 56\).

Answer:

\(56\)