Let's begin by looking at the given information and the desired outcome of the proof. Then we can write a paragraph proof!
&Given: m ∠ ADC = 120 and △ ABC is acute
&Prove: △ DBC is acute
From the diagram we can see that ∠ ADC and ∠ BDC form a linear pair. Recall that the Supplement Theorem tells us that if two angles form a linear pair, then they are supplementary. Therefore, by the Supplement Theorem, ∠ ADC and ∠ BDC are supplementary.
\begin{gathered}
\underline\textbf{Statement}\\
\angle ADC \text{ and } \angle BDC \text{ form a linear pair.} \\
\text{Therefore, by the Supplement Theorem, } \\ \angle ADC \text{ and } \angle BDC \text{ are supplementary.}
\end{gathered}
Next, by the definition of supplementary angles, we can conclude that the sum of the measures of ∠ ADC and ∠ BDC is 180.
\begin{gathered}
\underline\textbf{Statement}\\
\text{By the definition of supplementary angles, } \\
m \angle ADC+ m \angle BDC = 180.
\end{gathered}
We know that m ∠ ADC=120, so we can substitute 120 for m ∠ ADC in our equation.
\begin{gathered}
\underline\textbf{Statement}\\
\text{We know that } m \angle ADC=120, \\ \text{ so by substitution }
120+ m \angle BDC = 180.
\end{gathered}
Now, we can subtract 120 from both sides. This gives us m ∠ BDC = 60, and we can conclude that ∠ BDC is acute.
\begin{gathered}
\underline\textbf{Statement}\\
\text{Subtracting }120 \text{ from both sides, we get } \\
m \angle BDC=60. \text{ Therefore, } \angle BDC \text{ is acute.}
\end{gathered}
We are given that △ ABC is acute. By the definition of an acute triangle, we can conclude that ∠ B and ∠ C are acute.
\begin{gathered}
\underline\textbf{Statement}\\
\text{Since }\triangle ABC \text{ is acute, }
\angle B \text{ and } \angle C \text{ are acute. }
\end{gathered}
From the diagram we can tell that m ∠ C = m ∠ ACD + m ∠ BCD. Since ∠ C is acute and ∠ BCD is a part of this angle, we can conclude that ∠ BCD is also acute.
\begin{gathered}
\underline\textbf{Statement}\\
\text{We can tell that }m \angle C = m \angle ACD + m \angle BCD, \\ \text{ so } \angle BCD \text{ is also acute}.
\end{gathered}
Now, we know that ∠ BDC, ∠ B, and ∠ BCD are acute angles. Therefore, by the definition of an acute triangle, △ DBC is acute. This is what we wanted to prove!
\begin{gathered}
\underline\textbf{Statement}\\
\text{By the definition of an acute triangle, } \\ \triangle DBC \text{ is acute}.
\end{gathered}
Completed Proof
&Given: m ∠ ADC = 120 and △ ABC is acute
&Prove: △ DBC is acute
Proof: ∠ ADC and ∠ BDC form a linear pair. Therefore, by the Supplement Theorem, ∠ ADC and ∠ BDC are supplementary. By the definition of supplementary angles
m ∠ ADC + m ∠ BDC = 180.
We know that m ∠ ADC = 120, so by substitution 120+ m ∠ BDC = 180. Subtracting 120 from both sides, we get m ∠ BDC = 60. Therefore, ∠ BDC is acute. Since △ ABC is acute, ∠ B and ∠ C are acute. We can tell that m ∠ C= m ∠ ACD+ m ∠ BCD, so ∠ BCD is also acute. By the definition of an acute triangle, △ DBC is acute.
