Question

complete the paragraph to prove that \\( \triangle x y z \sim \triangle a b c \\).
start by applying a dilation to \\( \triangle x y z \\) centered at \\( y \\) with scale factor \\( \frac{a c}{x z} \\). this will map
\\( \triangle x y z \\) to \\( \triangle x^{prime} y z^{prime} \\). since the dilation is centered at \\( y, \overrightarrow{y x}=\overrightarrow{y x^{prime}} \\) and
. this means
will stay fixed. dilations preserve
, which
means
is congruent to \\( \angle a \\). the scale factor of the dilation is \\( \frac{a c}{x z} \\), so \\( \overline{x z} \\) is congruent
to
so, \\( \triangle x^{prime} y z^{prime} \\) is congruent to \\( \triangle a b c \\) by the
congruence theorem. by
the definition of congruence, there is a sequence of
that maps \\( \triangle x^{prime} y z^{prime} \\) to \\( \triangle a b c \\). this means there is a sequence of

Explanation:

Step1: Properties of dilation center

Since the dilation is centered at \(Y\), point \(Y\) will stay fixed.

Step2: Angle preservation in dilation

Dilations preserve angles. So, \(\angle XYZ\) (or \(\angle Y\) in \(\triangle XYZ\)) is congruent to \(\angle Y\) (or \(\angle X'YZ'\)) in \(\triangle X'YZ'\). Also, \(\angle X\) in \(\triangle XYZ\) is congruent to \(\angle A\) (after dilation and considering the similarity condition).

Step3: Side - length relationship

The scale factor of the dilation is \(\frac{AC}{XZ}\), so \(\overline{XZ}\) is congruent to \(\overline{AC}\) (after dilation, \(X'Z'=\frac{AC}{XZ}\times XZ = AC\)).

Step4: Congruence theorem

\(\triangle X'YZ'\) is congruent to \(\triangle ABC\) by the \(ASA\) (Angle - Side - Angle) congruence theorem. (We have \(\angle X'=\angle A\), \(X'Z' = AC\), and \(\angle Z'=\angle C\) (since angles are preserved in dilation and from the given angle - angle similarity condition)).

Step5: Similarity conclusion

By the definition of congruence, there is a sequence of rigid motions (translation, rotation, reflection) that maps \(\triangle X'YZ'\) to \(\triangle ABC\). This means there is a sequence of a dilation and rigid motions that maps \(\triangle XYZ\) to \(\triangle ABC\), so \(\triangle XYZ\sim\triangle ABC\)

Answer:

Start by applying a dilation to \(\triangle XYZ\) centered at \(Y\) with scale factor \(\frac{AC}{XZ}\). This will map \(\triangle XYZ\) to \(\triangle X'YZ'\). Since the dilation is centered at \(Y\), \(\overrightarrow{YX}=\overrightarrow{YX'}\) and \(Y\) will stay fixed. Dilations preserve angles, which means \(\angle X\) is congruent to \(\angle A\). The scale factor of the dilation is \(\frac{AC}{XZ}\), so \(\overline{XZ}\) is congruent to \(\overline{AC}\). So, \(\triangle X'YZ'\) is congruent to \(\triangle ABC\) by the \(ASA\) congruence theorem. By the definition of congruence, there is a sequence of rigid motions that maps \(\triangle X'YZ'\) to \(\triangle ABC\). This means there is a sequence of a dilation and rigid motions that maps \(\triangle XYZ\) to \(\triangle ABC\) (so \(\triangle XYZ\sim\triangle ABC\))