Question

prove the following statement.
if $\triangle abc \sim \triangle def$ and $\triangle def \sim \triangle ghk$, then $\triangle abc \sim \triangle ghk$.
two polygons are similar polygons if corresponding angles are \boxed{} and if the lengths of corresponding sides are \boxed{}
options for the first box: congruent, complementary, proportional, supplementary
options for the second box: (not fully visible, but related to similar polygons side condition)

Explanation:

Step1: Define similar polygon criteria

By definition, two polygons are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal (proportional).

Step2: Match to the blanks

First blank: Corresponding angles must be congruent.
Second blank: Corresponding sides must be proportional.

Step3: Prove triangle similarity transitivity

Sub-step3.1: Use similarity definition for $\Delta ABC \sim \Delta DEF$

$\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$, and $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=k_1$ where $k_1$ is a positive real number.

Sub-step3.2: Use similarity definition for $\Delta DEF \sim \Delta GHK$

$\angle D = \angle G$, $\angle E = \angle H$, $\angle F = \angle K$, and $\frac{DE}{GH}=\frac{EF}{HK}=\frac{DF}{GK}=k_2$ where $k_2$ is a positive real number.

Sub-step3.3: Transitive property for angles

$\angle A = \angle G$, $\angle B = \angle H$, $\angle C = \angle K$ (by transitivity of congruence).

Sub-step3.4: Transitive property for side ratios

$\frac{AB}{GH}=\frac{AB}{DE} \cdot \frac{DE}{GH}=k_1 \cdot k_2$, $\frac{BC}{HK}=\frac{BC}{EF} \cdot \frac{EF}{HK}=k_1 \cdot k_2$, $\frac{AC}{GK}=\frac{AC}{DF} \cdot \frac{DF}{GK}=k_1 \cdot k_2$. So $\frac{AB}{GH}=\frac{BC}{HK}=\frac{AC}{GK}$.

Sub-step3.5: Conclude similarity

By the definition of similar polygons, $\Delta ABC \sim \Delta GHK$.

Answer:

1. First dropdown: congruent
2. Second dropdown: proportional
3. The proof confirms $\Delta ABC \sim \Delta GHK$ as required.