1. **Problem Statement:** Given triangle $\triangle PQR$ with point $S$ on side $PQ$ such that $PS$ is perpendicular to $SR$ and $RS$ is perpendicular to $QR$, complete the table of geometric means for segments. 2. **Understanding Geometric Mean in Right Triangles:** In right triangles, the altitude to the hypotenuse creates two smaller right triangles similar to the original. The altitude length is the geometric mean of the two segments it divides the hypotenuse into. 3. **Applying the Geometric Mean Theorem:** - For altitude $PS$ to hypotenuse $QR$, $PS = \sqrt{QS \times PR}$. - For altitude $QS$ to hypotenuse $PR$, $QS = \sqrt{PS \times PQ}$. 4. **Completing the Table:** | Segments | Geometric Mean | |-------------------|----------------| | $PS$ and $QS$ | $PR$ | | $PQ$ and $QS$ | $PR$ | 5. **Similarity of Triangles $\triangle DEF$ and $\triangle EFG$:** - $\triangle EFG$ shares angle $F$ with $\triangle DEF$ and has a right angle at $E$. - By AA similarity (two angles equal), $\triangle EFG \sim \triangle DEF$. 6. **Answer to Similarity Question:** - The triangle similar to $\triangle DEF$ is $\triangle EFG$ (Option C). 7. **Proportion to Find $a$ in Right Triangle with legs 16 and $a$, hypotenuse 18:** - Using Pythagorean theorem: $16^2 + a^2 = 18^2$. - The proportion from the options that matches the geometric mean relation is $\frac{18}{a} = \frac{a}{16}$ (Option D). **Final answers:** - Geometric mean table completed as above. - Similar triangle: $\triangle EFG$. - Proportion to find $a$: $\frac{18}{a} = \frac{a}{16}$.