Question

triangle def was dilated according to the rule ( d_{o,\frac{1}{3}}(x,y)
ightarrow(\frac{1}{3}x,\frac{1}{3}y) ) to create similar triangle ( def ).
which statements are true? select three options.
( squareangle f ) corresponds to ( angle f ).
( square ) segment ( ee ) is parallel to segment ( ff ).
( square ) the distance from point ( d ) to the origin is ( \frac{1}{3} ) the distance of point ( d ) to the origin.
( square ) the measure of ( angle e ) is ( \frac{1}{3} ) the measure of ( angle e ).
( square\triangle defsim\triangle def )

Explanation:

Step1: Properties of dilation

Dilation is a transformation that changes the size of a figure but not its shape. Corresponding angles of similar (dilated) figures are congruent. So, \(\angle F\) in \(\triangle DEF\) and \(\angle F'\) in \(\triangle D'E'F'\) are congruent (they correspond).

Step2: Dilation and distance from the center

If a point \(P(x,y)\) is dilated about the origin \(O\) with a scale factor \(k=\frac{1}{3}\) according to the rule \(D_{O,k}(x,y)=(\frac{1}{3}x,\frac{1}{3}y)\), the distance from a point \(P(x,y)\) to the origin \(d=\sqrt{x^{2}+y^{2}}\), and for the dilated point \(P'( \frac{1}{3}x,\frac{1}{3}y)\), the distance \(d'=\sqrt{(\frac{1}{3}x)^{2}+(\frac{1}{3}y)^{2}}=\frac{1}{3}\sqrt{x^{2}+y^{2}}\). So, the distance from point \(D'\) to the origin is \(\frac{1}{3}\) the distance of point \(D\) to the origin.

Step3: Similarity of triangles

By the definition of dilation, if \(\triangle DEF\) is dilated to get \(\triangle D'E'F'\) with a scale factor \(k = \frac{1}{3}\), then \(\triangle DEF\sim\triangle D'E'F'\) (all corresponding angles are equal and corresponding sides are in proportion).

For the statement “Segment \(EE'\) is parallel to segment \(FF'\)”:
Let \(E(x_1,y_1)\) and \(E'(\frac{1}{3}x_1,\frac{1}{3}y_1)\), \(F(x_2,y_2)\) and \(F'(\frac{1}{3}x_2,\frac{1}{3}y_2)\). The slope of \(EE'\) is \(m_{EE'}=\frac{\frac{1}{3}y_1 - y_1}{\frac{1}{3}x_1 - x_1}=\frac{-\frac{2}{3}y_1}{-\frac{2}{3}x_1}=\frac{y_1}{x_1}\) (if \(x_1
eq0\)), and the slope of \(FF'\) is \(m_{FF'}=\frac{\frac{1}{3}y_2 - y_2}{\frac{1}{3}x_2 - x_2}=\frac{-\frac{2}{3}y_2}{-\frac{2}{3}x_2}=\frac{y_2}{x_2}\). Since \(E\) and \(F\) are two different points of \(\triangle DEF\), \(\frac{y_1}{x_1}
eq\frac{y_2}{x_2}\) in general, so \(EE'\) and \(FF'\) are not parallel.

For the statement “The measure of \(\angle E'\) is \(\frac{1}{3}\) the measure of \(\angle E\)”:
Since \(\triangle DEF\sim\triangle D'E'F'\), \(\angle E=\angle E'\) (corresponding angles of similar triangles are congruent), not \(\angle E'=\frac{1}{3}\angle E\).

Answer:

• ∠F corresponds to ∠F′.
• The distance from point D′ to the origin is \(\frac{1}{3}\) the distance of point D to the origin.
• \(\triangle DEF\sim\triangle D'E'F'\)