Problem Explanation:

In this problem, $\overline{AG}$ is an angle bisector in $\triangle ABC$. This means that $m\angle BAG$ and $m\angle CAG$ are equal in measure, and their sum will give $m\angle CAB$.

The given expressions are:

• $m\angle BAG = 4x + 5$
• $m\angle CAG = 2x + 15$

Steps to Solve:

1. Since $\overline{AG}$ bisects $\angle CAB$: $m\angle BAG = m\angle CAG$ Therefore: $4x + 5 = 2x + 15$

2. Solve for $x$: Subtract $2x$ from both sides: $2x + 5 = 15$ Subtract $5$ from both sides: $2x = 10$ Divide by 2: $x = 5$

3. Substitute $x = 5$ back into the expressions for $m\angle BAG$ and $m\angle CAG$:

   • $m\angle BAG = 4x + 5 = 4(5) + 5 = 20 + 5 = 25^\circ$
   • $m\angle CAG = 2x + 15 = 2(5) + 15 = 10 + 15 = 25^\circ$

4. Since $\overline{AG}$ bisects $\angle CAB$: $m\angle CAB = m\angle BAG + m\angle CAG = 25^\circ + 25^\circ = 50^\circ$

Final Answer:

• $x = 5$
• $m\angle BAG = 25^\circ$
• $m\angle CAB = 50^\circ$

Would you like further clarification or a detailed breakdown of any part?

Here are 5 related questions to expand your understanding:

1. What are the properties of angle bisectors in a triangle?
2. Can an angle bisector divide opposite sides proportionally in all triangles?
3. How does this concept extend to finding unknown variables in other geometric figures?
4. How is the Angle Bisector Theorem used differently from this direct angle equality method?
5. Can $\overline{AG}$ still be called an angle bisector if the triangle is not isosceles?

Tip: Always verify your final values by substituting back into the original equations to ensure consistency.