Solution
- The triangles ABD and BDC share a common side, BD and have sides AB and BC equal.
- Since two sides from both are equal, then, we can conclude that their third sides are equal as well.
- Thus, we can equate the expression for the sides AD and DC
- We have:
[tex]\begin{gathered} AD=DC \\ 9x-15=7x-1 \\ \text{ Subtract }7x\text{ from both sides, Add 15 to both sides} \\ 9x-7x=15-1 \\ 2x=14 \\ \text{ Divide both sides by 2} \\ x=\frac{14}{2} \\ \\ x=7 \\ \\ AC=AD+DC \\ AC=9x-15+(7x-1) \\ AC=9x+7x-16 \\ AC=16x-16=16(x-1) \\ \text{ But we now know that }x=7 \\ AC=16(7-1)=16\times6 \\ \\ AC=96 \end{gathered}[/tex]
Final Answer
x = 7
AC = 96
