Let O(0, 0), A(-1, 2) and B(1, 3) be the vert | vedclass

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(C) Step $1$: Calculate the lengths of sides $OA$ and $OB$.$OA = \sqrt{(-1-0)^2 + (2-0)^2} = \sqrt{1+4} = \sqrt{5}$.$OB = \sqrt{(1-0)^2 + (3-0)^2} = \

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$5\sqrt{4-2\sqrt{2}}$

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(C) Step $1$: Calculate the lengths of sides $OA$ and $OB$.$OA = \sqrt{(-1-0)^2 + (2-0)^2} = \sqrt{1+4} = \sqrt{5}$.$OB = \sqrt{(1-0)^2 + (3-0)^2} = \sqrt{1+9} = \sqrt{10}$.Step $2$: Use the Angle Bisector Theorem. Point $D$ divides $AB$ in the ratio $OA:OB = \sqrt{5}:\sqrt{10} = 1:\sqrt{2}$.Step $3$: Find the coordinates of $D$ using the section formula:$D = \left( \frac{1(1) + \sqrt{2}(-1)}{1+\sqrt{2}}, \frac{1(3) + \sqrt{2}(2)}{1+\sqrt{2}} \right) = \left( \frac{1-\sqrt{2}}{1+\sqrt{2}}, \frac{3+2\sqrt{2}}{1+\sqrt{2}} \right)$.Step $4$: Calculate length $OD = \sqrt{\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}\right)^2 + \left(\frac{3+2\sqrt{2}}{1+\sqrt{2}}\right)^2} = \frac{\sqrt{(1-\sqrt{2})^2 + (3+2\sqrt{2})^2}}{1+\sqrt{2}} = \frac{\sqrt{(1+2-2\sqrt{2}) + (9+8+12\sqrt{2})}}{1+\sqrt{2}} = \frac{\sqrt{20+10\sqrt{2}}}{1+\sqrt{2}}$.Step $5$: Calculate length $AB = \sqrt{(1 - (-1))^2 + (3-2)^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$.Step $6$: $OD \cdot AB = \frac{\sqrt{20+10\sqrt{2}}}{1+\sqrt{2}} \cdot \sqrt{5} = \frac{\sqrt{100+50\sqrt{2}}}{1+\sqrt{2}}$.To simplify, we square the expression: $\left( \frac{\sqrt{100+50\sqrt{2}}}{1+\sqrt{2}} \right)^2 = \frac{100+50\sqrt{2}}{3+2\sqrt{2}} = \frac{50(2+\sqrt{2})(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})} = 50(6 - 4\sqrt{2} + 3\sqrt{2} - 4) = 50(2-\sqrt{2}) = 100 - 50\sqrt{2}$.Thus, $OD \cdot AB = \sqrt{100-50\sqrt{2}} = 5\sqrt{4-2\sqrt{2}}$.

Let $O(0, 0)$, $A(-1, 2)$ and $B(1, 3)$ be the vertices of $\Delta OAB$. The bisector of $\angle O$ intersects side $AB$ at point $D$. The value of $OD \cdot AB$ is equal to...

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