Angles formed by the segment [tex]\overline{XZ}[/tex] in the triangles ΔWXZ, and ΔXYZ, are equal and the given corresponding sides are proportional.
- The option that best completes the proof showing that ΔWXZ ~ ΔXYZ is; 16 over 12 equals 12 over 9
Reasons:
The proof showing that ΔWXZ ~ ΔXYZ is presented as follows;
Segment [tex]\overline{XZ}[/tex] is perpendicular to segment [tex]\overline{WY}[/tex]
∠WZX and ∠XZY are right angles by definition of [tex]\overline{XZ}[/tex] perpendicular to [tex]\overline{WY}[/tex]
∠WZX in ΔWXZ = ∠XZY in ΔXYZ = 90° (definition)
[tex]\displaystyle \frac{WZ}{XZ} = \frac{16}{12} = \mathbf{ \frac{4}{3}}[/tex]
[tex]\displaystyle \mathbf{ \frac{XZ}{ZY}} = \frac{12}{9} = \frac{4}{3}[/tex]
Therefore;
- [tex]\displaystyle \frac{16}{12} = \frac{12}{9}[/tex], which gives, [tex]\displaystyle \mathbf{\frac{WZ}{XZ} }= \frac{XZ}{ZY}[/tex]
Given that two sides of ΔWXZ are proportional to two sides of ΔXYZ, and
that the included angles between the two sides, ∠WZX and ∠XZY are
congruent, the two triangles, ΔWXZ and ΔXYZ are similar by Side-Angle-
Side, SAS, similarity postulate.
The option that best completes the proof is therefore;
- [tex]\displaystyle \frac{16}{12} = \frac{12}{9}[/tex] which is; 16 over 12 equals 12 over 9
Learn more about the SAS similarity postulate here:
https://brainly.com/question/11923416
