1. **Problem statement:** We have two right-angled triangles ABC and CDE touching at point C, with an equilateral triangle ACE formed by connecting points A, C, and E. Given AB = 2.5/10 = 0.25 units, we need to find lengths AB, BC, CD, DE and the angle at CED. 2. **Given:** - Triangle ABC is right-angled at C. - Triangle CDE is right-angled at D. - Triangle ACE is equilateral, so all sides AC = CE = AE and all angles 60°. - AB = 0.25 units. 3. **Find AB:** Given directly as $AB = \frac{2.5}{10} = 0.25$ units. 4. **Find BC:** In right triangle ABC, by Pythagoras theorem: $$AB^2 = AC^2 + BC^2$$ Since ACE is equilateral, $AC = AE = CE$. We know $AB = 0.25$ and $AC = CE$. 5. **Find AC:** Since ACE is equilateral, all sides equal, so $AC = AE = CE$. We need to find $AC$ to find $BC$. 6. **Express BC in terms of AC:** From triangle ABC: $$BC = \sqrt{AB^2 - AC^2} = \sqrt{0.25^2 - AC^2}$$ 7. **Find CD and DE:** Triangle CDE is right angled at D. Given angle at FCD is twice angle at CDE, and F is on horizontal extension of angle ACB. 8. **Find angle CED:** Since ACE is equilateral, angle ACE = 60°. Given angle FCD = 2 * angle CDE. Since F lies on horizontal extension of angle ACB, angle FCD + angle CDE + angle CED = 180° (straight line). Let angle CDE = x, then angle FCD = 2x. So: $$2x + x + \angle CED = 180^\circ \Rightarrow 3x + \angle CED = 180^\circ$$ 9. **Find angle CED:** In triangle CDE, sum of angles = 180°: $$\angle CDE + \angle CED + 90^\circ = 180^\circ \Rightarrow x + \angle CED = 90^\circ$$ From step 8: $$3x + \angle CED = 180^\circ$$ Subtracting the two equations: $$(3x + \angle CED) - (x + \angle CED) = 180^\circ - 90^\circ \Rightarrow 2x = 90^\circ \Rightarrow x = 45^\circ$$ 10. **Calculate angle CED:** From step 9: $$x + \angle CED = 90^\circ \Rightarrow 45^\circ + \angle CED = 90^\circ \Rightarrow \angle CED = 45^\circ$$ 11. **Find CD and DE:** In right triangle CDE with right angle at D, and angles 45° at CDE and 45° at CED, triangle CDE is isosceles right triangle. So: $$CD = DE$$ 12. **Find lengths CD and DE:** Since ACE is equilateral, $AC = CE$. From triangle ACE, $AE = AC$. Since $AE = AC$, and $CE = AC$, and $CE$ is side of triangle CDE. 13. **Summary of known lengths:** - $AB = 0.25$ - $AC = CE = AE$ - Triangle CDE is isosceles right angled at D with $CD = DE$ 14. **Find BC:** From triangle ABC: $$BC = \sqrt{AB^2 - AC^2} = \sqrt{0.25^2 - AC^2}$$ 15. **Find CD and DE:** Since $CE = AC$, and triangle CDE is right angled at D with $CD = DE$, by Pythagoras: $$CE^2 = CD^2 + DE^2 = 2 CD^2 \Rightarrow CD = DE = \frac{CE}{\sqrt{2}} = \frac{AC}{\sqrt{2}}$$ **Final answers:** - $AB = 0.25$ - $BC = \sqrt{0.25^2 - AC^2}$ - $\angle CED = 45^\circ$ - $CD = DE = \frac{AC}{\sqrt{2}}$ Since $AC$ is not given numerically, lengths $BC$, $CD$, and $DE$ are expressed in terms of $AC$.