In the figure below
1) Using the theorem of similar triangles (ΔBXY and ΔBAC),
[tex]\frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC}[/tex]
Where
[tex]\begin{gathered} BX=4 \\ BA=5 \\ BY=6 \\ BC\text{ = x} \end{gathered}[/tex]
Thus,
[tex]\begin{gathered} \frac{4}{5}=\frac{6}{x} \\ \text{cross}-\text{multiply} \\ 4\times x=6\times5 \\ 4x=30 \\ \text{divide both sides by the coefficient of x, which is 4} \\ \text{thus,} \\ \frac{4x}{4}=\frac{30}{4} \\ x=7.5 \end{gathered}[/tex]
thus, BC = 7.5
2) BX = 9, BA = 15, BY = 15, YC = y
In the above diagram,
[tex]\begin{gathered} BC=BY+YC \\ \Rightarrow BC=15\text{ + y} \end{gathered}[/tex]
Thus, from the theorem of similar triangles,
[tex]\begin{gathered} \frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC} \\ \frac{9}{15}=\frac{15}{15+y} \end{gathered}[/tex]
solving for y, we have
[tex]\begin{gathered} \frac{9}{15}=\frac{15}{15+y} \\ \text{cross}-\text{multiply} \\ 9(15+y)=15(15) \\ \text{open brackets} \\ 135+9y=225 \\ \text{collect like terms} \\ 9y\text{ = 225}-135 \\ 9y=90 \\ \text{divide both sides by the coefficient of y, which is 9} \\ \text{thus,} \\ \frac{9y}{9}=\frac{90}{9} \\ \Rightarrow y=10 \end{gathered}[/tex]
thus, YC = 10.
