1. **Problem statement:** We are given triangles BOX and DOP with right angles at $\angle BXO$ and $\angle ODP$. Point O lies between X and D on segment XD. Ray OL bisects $\angle BOP$. Given $OP=10$, $BO=12$, and $XD=13$, find the length of $OX$. 2. **Key observations and formulas:** - Since $\angle BXO = 90^\circ$ and $\angle ODP = 90^\circ$, triangles BOX and DOP are right triangles. - Point O lies on segment XD, so $XD = XO + OD = 13$. - Ray OL bisects $\angle BOP$, so by the Angle Bisector Theorem in triangle BOP, $\frac{BL}{LP} = \frac{BO}{OP} = \frac{12}{10} = \frac{6}{5}$. 3. **Find length OD:** - Since $\angle ODP = 90^\circ$, triangle DOP is right angled at D. - Using Pythagoras theorem in triangle DOP: $$DO^2 + OP^2 = DP^2$$ - We don't know DP yet, but we can find DP using the fact that B and P lie on the same half-plane relative to line XD and OL bisects $\angle BOP$. 4. **Find length BX:** - Triangle BOX is right angled at X, so: $$BX^2 = BO^2 + OX^2 = 12^2 + OX^2 = 144 + OX^2$$ 5. **Use the fact that OL is perpendicular to DX:** - Since $OL \perp DX$, and O lies on XD, OL is the altitude from O to line segment DX. 6. **Use the segment addition on XD:** - $XD = XO + OD = 13$ - We want to find $OX$, so if we find $OD$, we can find $OX = 13 - OD$. 7. **Use similarity or coordinate geometry approach:** - Place points on coordinate axis for clarity: - Let X be at $0,0$ - Let D be at $13,0$ - Then O lies on the x-axis between 0 and 13, at $(x,0)$ where $0 < x < 13$ 8. **Coordinates of points B and P:** - Since $\angle BXO = 90^\circ$, and B is on the same half-plane as P relative to XD (x-axis), B and P have positive y-coordinates. - Point B lies such that $BO = 12$, and B is perpendicular to X at O. - So B is at $(0,12)$ because $BO = 12$ and O is at $(x,0)$, so distance from B to O is 12: $$BO = \sqrt{(0 - x)^2 + (12 - 0)^2} = 12$$ - Simplify: $$\sqrt{(x)^2 + 144} = 12 \Rightarrow x^2 + 144 = 144 \Rightarrow x^2 = 0 \Rightarrow x=0$$ - Contradiction since O is between X(0,0) and D(13,0), so O cannot be at 0. 9. **Reconsider B coordinates:** - Since B is on the same half-plane as P, and $\angle BXO = 90^\circ$, B lies on a line perpendicular to XD at X. - X is at (0,0), so line perpendicular to XD at X is vertical line x=0. - So B is at (0,b) with $b>0$. - Distance BO is 12: $$BO = \sqrt{(0 - x)^2 + (b - 0)^2} = 12$$ - So: $$\sqrt{x^2 + b^2} = 12 \Rightarrow x^2 + b^2 = 144$$ 10. **Similarly, P lies on perpendicular to XD at D (13,0):** - P is at (13,p) with $p>0$. - Distance OP is 10: $$OP = \sqrt{(13 - x)^2 + (p - 0)^2} = 10$$ - So: $$(13 - x)^2 + p^2 = 100$$ 11. **Since OL bisects $\angle BOP$, the coordinates of L satisfy the angle bisector condition:** - Vector OB = (0 - x, b - 0) = (-x, b) - Vector OP = (13 - x, p - 0) = (13 - x, p) - The angle bisector direction vector OL is proportional to the sum of the unit vectors in directions OB and OP: $$\vec{OL} \propto \frac{\vec{OB}}{|OB|} + \frac{\vec{OP}}{|OP|}$$ - Since $OL \perp XD$, OL is vertical, so its x-component is zero: $$\frac{-x}{12} + \frac{13 - x}{10} = 0$$ 12. **Solve for x:** $$-\frac{x}{12} + \frac{13 - x}{10} = 0$$ Multiply both sides by 60 (LCM of 12 and 10): $$-5x + 6(13 - x) = 0$$ $$-5x + 78 - 6x = 0$$ $$-11x + 78 = 0$$ $$11x = 78$$ $$x = \frac{78}{11} = 7.09$$ 13. **Find b:** $$x^2 + b^2 = 144$$ $$7.09^2 + b^2 = 144$$ $$50.28 + b^2 = 144$$ $$b^2 = 93.72$$ $$b = \sqrt{93.72} = 9.68$$ 14. **Find p:** $$(13 - x)^2 + p^2 = 100$$ $$(13 - 7.09)^2 + p^2 = 100$$ $$5.91^2 + p^2 = 100$$ $$34.92 + p^2 = 100$$ $$p^2 = 65.08$$ $$p = \sqrt{65.08} = 8.07$$ 15. **Check OL perpendicular to XD:** - OL vector is sum of unit vectors: $$\vec{OL} = \left(\frac{-x}{12} + \frac{13 - x}{10}, \frac{b}{12} + \frac{p}{10}\right) = (0, y)$$ - We already confirmed x-component is zero. 16. **Find OX:** - Since O is at $x=7.09$, and X is at 0, $$OX = 7.09$$ **Final answer:** $$\boxed{7.09}$$ units (rounded to nearest hundredth).