Triangle Construction Aebc61

1. **Problem Statement:** We need to construct $\triangle ABC$ where $AB = 5cm$, $DC = 5cm$, and $D$ is the midpoint of $AB$. We want to find the lengths of $AC$ and $BC$. 2. **Understanding the problem:** Since $D$ is the midpoint of $AB$, $AD = DB = \frac{AB}{2} = \frac{5cm}{2} = 2.5cm$. 3. **Using the compass construction:** $DC = 5cm$ means the segment from $D$ to $C$ is $5cm$. Since $D$ lies on $AB$, and $C$ is a point such that $DC = 5cm$, $C$ lies on a circle centered at $D$ with radius $5cm$. 4. **Finding $AC$ and $BC$:** $AC$ and $BC$ are the distances from $A$ and $B$ to $C$ respectively. 5. **Using the Pythagorean theorem:** Since $D$ is midpoint of $AB$, $AB$ is horizontal, and $C$ lies vertically above or below $D$ at distance $5cm$, $\triangle ADC$ and $\triangle BDC$ are right triangles with legs $AD = 2.5cm$ and $DC = 5cm$. 6. Calculate $AC$: $$AC = \sqrt{AD^2 + DC^2} = \sqrt{(2.5)^2 + 5^2} = \sqrt{6.25 + 25} = \sqrt{31.25}$$ 7. Calculate $BC$: Similarly, $BC = \sqrt{DB^2 + DC^2} = \sqrt{(2.5)^2 + 5^2} = \sqrt{31.25}$ 8. Simplify $\sqrt{31.25}$: $$\sqrt{31.25} = \sqrt{\frac{125}{4}} = \frac{\sqrt{125}}{2} = \frac{5\sqrt{5}}{2} \approx 5.59cm$$ **Final answer:** $$AC = BC = \frac{5\sqrt{5}}{2} \approx 5.59cm$$