$\large{YES}\ \dfrac{BC}{YZ}=\dfrac{AC}{XY}=\dfrac{AB}{XZ}=\dfrac{2}{1}\\\\\text{and}\ \angle B\cong\angle Z,\ \angle C\cong\angle Y,\ \angle A\cong\angle X$

Step-by-step explanation:

$BC\to ZY\\AB\to XZ\\AC\to XY\\\angle A\to\angle X\\\angle B\to\angle Z\\\angle C\to\angle Y$

$\text{We have:}\\\\\triangle ABC:\ AB=20,\ BC=12,\ AC=18\\\triangle XZY:\ XZ=10,\ YZ=6,\ XY=9\\\\\text{check the ratio:}\\\\\dfrac{AB}{XZ}=\dfrac{20}{10}=2\\\\\dfrac{BC}{YZ}=\dfrac{12}{6}=2\\\\\dfrac{AC}{XY}=\dfrac{18}{9}=2\\\\\text{CORRECT :)}\\\\\angle B\cong\angle Z,\ \angle C\cong\angle Y\\\\\text{We know that: The sum of the acute angles in a right triangle is}\ 90^o.\\\\m\angle B+m\angle A=90^o\ \text{and}\ m\angle Z+m\angle X=90^o$

$\text{We know}\ m\angle B=m\angle Z\to \ m\angle A=m\angle X.\\\text{therefore}\ \angle A\cong\angle X$

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3 years ago