1. **State the problem:** Prove the Boolean identity $AC + BC = (A + B)C$ using a truth table. 2. **Understand the expressions:** The expressions $AC + BC$ and $(A + B)C$ represent Boolean OR and AND operations. We want to show they produce the same outputs for all values of $A$, $B$, and $C$. 3. **Analyze each row of the truth table:** For each triple $(A, B, C)$, compute $AC$, $BC$, $AC + BC$, $A + B$, and $(A + B)C$. 4. **Check equivalence in the table:** According to the provided truth table: - For each row, the value under $AC + BC$ matches the value under $(A + B)C$. - For example, row 2: $A=0$, $B=0$, $C=1$ - $AC = 0 imes 1 = 0$ - $BC = 0 imes 1 = 0$ - $AC + BC = 0 + 0 = 0$ - $A + B = 0 + 0 = 0$ - $(A + B)C = 0 imes 1 = 0$ Both sides equal 0. - Row 6: $A=1$, $B=0$, $C=1$ - $AC = 1 imes 1 = 1$ - $BC = 0 imes 1 = 0$ - $AC + BC = 1 + 0 = 1$ - $A + B = 1 + 0 = 1$ - $(A + B)C = 1 imes 1 = 1$ Both sides equal 1. 5. **Conclude:** Because $AC + BC$ and $(A + B)C$ have identical outputs for all combinations of inputs $A$, $B$, and $C$, the Boolean identity is proven. **Final answer:** $$AC + BC = (A + B)C$$