1. **Problem 1: Prove that $3 - 2\sqrt{5}$ is irrational given $\sqrt{5}$ is irrational.** 2. We start by assuming the contrary: suppose $3 - 2\sqrt{5}$ is rational. 3. Let $x = 3 - 2\sqrt{5}$ and assume $x$ is rational. 4. Rearranging, we get $2\sqrt{5} = 3 - x$. 5. Dividing both sides by 2, $\sqrt{5} = \frac{3 - x}{2}$. 6. Since $x$ is rational, $3 - x$ is rational, and dividing by 2 keeps it rational. 7. This implies $\sqrt{5}$ is rational, which contradicts the given fact that $\sqrt{5}$ is irrational. 8. Therefore, our assumption is false, and $3 - 2\sqrt{5}$ is irrational. 1. **Problem 2: In triangle $\triangle PQR$, segment $ST$ is drawn such that $\angle PRQ = \angle STQ$ and $ST$ divides $QR$ in the ratio $2:3$. Given $PR = 20$ cm, find the length of $ST$.** 2. Since $ST$ divides $QR$ in ratio $2:3$, let $QS = 2k$ and $SR = 3k$, so $QR = 5k$. 3. The condition $\angle PRQ = \angle STQ$ implies triangles $PRQ$ and $STQ$ are similar by the Angle-Angle (AA) similarity criterion. 4. From similarity, corresponding sides are proportional: $$\frac{ST}{PR} = \frac{TQ}{RQ}$$ 5. Since $ST$ divides $QR$ at $S$ and $T$, and $ST$ divides $QR$ in ratio $2:3$, point $T$ lies on $QR$ such that $QT:TR = 2:3$. 6. Therefore, $TQ = 2k$ and $RQ = 5k$. 7. Substitute into the proportion: $$\frac{ST}{20} = \frac{2k}{5k} = \frac{2}{5}$$ 8. Solve for $ST$: $$ST = 20 \times \frac{2}{5} = 8$$ 9. Hence, the length of $ST$ is $8$ cm.