Lines $L_1 : y - x = 0$ and $L_2 : 2x + y = 0$ intersect the line $L_3 : y + 2 = 0$ at $P$ and $Q$, respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersects $L_3$ at $R$. STATEMENT-I : The ratio $PR : RQ$ equals $2\sqrt{2} : \sqrt{5}$. STATEMENT-II : In any triangle, bisector of an angle divides the triangle into two similar triangles.

Lines L₁ : y - x = 0 and L₂ : 2x + y = 0 intersect the line L₃ : y + 2 = 0 at P and Q, respectively. The bisector of the acute angle between L₁ and L₂ intersects L₃ at R. STATEMENT-I : The ratio PR : RQ equals 2√{2} : √{5}. STATEMENT-II : In any triangle, bisector of an angle divides the triangle into two similar triangles.

$\text{Given: } L_1 : y - x = 0,\quad L_2 : 2x + y = 0,\quad L_3 : y + 2 = 0$

$P = L_1 \cap L_3,\qquad Q = L_2 \cap L_3$

$\text{To check: (I) } PR : RQ = 2\sqrt{2} : \sqrt{5}$

$\text{(II) angle bisector in a triangle gives two similar triangles}$

$L_1 : y = x$

$L_2 : y = -2x$

$L_3 : y + 2 = 0 \ \Rightarrow \ y = -2$

$L_1 \cap L_3:\ x = -2 \ \Rightarrow\ P(-2,-2)$

$L_2 \cap L_3:\ -2x = -2 \ \Rightarrow\ x = 1 \ \Rightarrow\ Q(1,-2)$

$OP \subset L_1,\quad OQ \subset L_2$

⇒ ∠ POQ = acute angle between L₁ and L₂

1. angle-bisector theorem in triangle OPQ

$2.\ \text{find the lengths } OP \text{ and } OQ$

$3.\ \text{check Statement-I, then Statement-II}$

$\frac{PR}{RQ} = \frac{OP}{OQ}\quad(\text{angle-bisector theorem})$

$OP = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$

$OQ = \sqrt{1^2 + (-2)^2} = \sqrt{5}$

$\frac{PR}{RQ} = \frac{2\sqrt{2}}{\sqrt{5}}$

$\Rightarrow\ PR : RQ = 2\sqrt{2} : \sqrt{5}\ \Rightarrow\ \text{Statement-I is TRUE}$

II claims triangle OPR ∼ triangle OQR

$\text{but the theorem only gives } \frac{PR}{RQ} = \frac{OP}{OQ}$

∠ ORP + ∠ ORQ = 180^∘ (P, R, Q collinear) similar ⇒ ∠ ORP = ∠ ORQ = 90^∘ ⇒ OR ⊥ PQ

$\text{here } OR \text{ is not perpendicular to } PQ$

$\Rightarrow\ \text{Statement-II is FALSE}$

$\boxed{\text{Statement-I True, Statement-II False} \ \Rightarrow\ (3)}$