Question

δxyz was reflected over a vertical line, then dilated by a scale factor of $\frac{1}{2}$, resulting in $δxyz$. which must be true of the two triangles? select three options. $δxyz sim δxyz$ $∠xzy = ∠yzx$ $overline{yx}=overline{yx}$ $xz = 2xz$ $m∠yxz = 2m∠yxz$

Explanation:

Step1: Properties of Reflection and Dilation

Reflection preserves shape and size (is a rigid transformation), dilation with scale factor \(k=\frac{1}{2}\) changes the size. For two triangles, if one is obtained from the other by a sequence of rigid - transformations (reflection is a rigid transformation) and dilation, they are similar. So, \(\triangle XYZ\sim\triangle X'Y'Z'\) (by the definition of similar triangles: if there is a sequence of rigid - motions and dilations that maps one triangle to the other).

Step2: Angle - Preservation in Similar Triangles

Since \(\triangle XYZ\sim\triangle X'Y'Z'\), corresponding angles are equal. \(\angle XZY\) and \(\angle Y'Z'X'\) are not corresponding angles. But for similar triangles \(\triangle XYZ\sim\triangle X'Y'Z'\), \(\angle XZY=\angle X'Z'Y'\) (corresponding angles). However, for the side - length relationship, if the scale factor of dilation is \(k = \frac{1}{2}\), then if we consider the pre - image side \(a\) (from \(\triangle XYZ\)) and the image side \(a'\) (from \(\triangle X'Y'Z'\)), \(a'=k\times a\). Let \(a = XZ\) and \(a'=X'Z'\), then \(X'Z'=\frac{1}{2}XZ\), or \(XZ = 2X'Z'\). Also, angles in similar triangles are equal, \(m\angle YXZ=m\angle Y'X'Z'\) (not \(m\angle YXZ = 2m\angle Y'X'Z'\)). And \(\overline{YX}
eq\overline{Y'X'}\) because of the dilation (scale factor \(\frac{1}{2}\)).

Answer:

\(\triangle XYZ\sim\triangle X'Y'Z'\), \(XZ = 2X'Z'\)