Question

In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find $\frac{\mathrm{AX}}{\mathrm{XY}}$.

Answer 1

In △ABY, ∠YBX = ∠XBA
$\frac{\mathrm{AX}}{\mathrm{XY}}=\frac{\mathrm{AB}}{\mathrm{BY}}...\left(\mathrm{I}\right)\left(\mathrm{By}\mathrm{angle}\mathrm{bisector}\mathrm{theorem}\right)$
In △ACY, ∠YCX = ∠XCA
​$\frac{\mathrm{AX}}{\mathrm{XY}}=\frac{\mathrm{AC}}{\mathrm{CY}}...\left(\mathrm{II}\right)\left(\mathrm{By}\mathrm{angle}\mathrm{bisector}\mathrm{theorem}\right)$
From (I) and (II)
$$\begin{gathered} \frac{\mathrm{AC}}{\mathrm{CY}}=\frac{\mathrm{AB}}{\mathrm{BY}} \\ \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}-\mathrm{BY}}=\frac{\mathrm{AB}}{\mathrm{BY}} \\ \Rightarrow \frac{4}{6-\mathrm{BY}}=\frac{5}{\mathrm{BY}} \end{gathered}$$
$$\begin{gathered} \Rightarrow 4\mathrm{BY}=30-5\mathrm{BY} \\ \Rightarrow 9\mathrm{BY}=30 \\ \Rightarrow \mathrm{BY}=\frac{10}{3} \end{gathered}$$
From (I), we have
​$$\begin{gathered} \frac{\mathrm{AX}}{\mathrm{XY}}=\frac{5}{{\displaystyle \frac{10}{3}}} \\ =\frac{3}{2} \end{gathered}$$