Question

In the given figure, XY ∥ AC and XY divides ∆ABC into two regions, equal in area. Show that $\frac{AX}{AB}=\frac{(2-\sqrt{2})}{2}.$

Answer 1

$$\begin{gathered} \mathrm{In}△ABC\mathrm{and}△BXY,\mathrm{we}\mathrm{have}: \\ \angle B=\angle B \\ \angle BXY=\angle BAC(\mathrm{Corresponding}\mathrm{angles}) \\ \mathrm{Thus},△ABC~△BXY(AAcriterion) \\ \therefore \frac{ar(△ABC)}{ar(△BXY)}=\frac{A{B}^2}{B{X}^2}=\frac{A{B}^2}{{\left(AB-AX\right)}^2}...\left(\mathrm{i}\right) \\ \mathrm{Also},\frac{ar(△ABC)}{ar(△BXY)}=\frac{2}{1}\{\because ar(△BXY)=ar(trapeziumAXYC)\}...\left(\mathrm{ii}\right) \\ \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{have}: \\ \frac{A{B}^2}{{\left(AB-AX\right)}^2}=\frac{2}{1} \\ \Rightarrow \frac{AB}{\left(AB-AX\right)}=\sqrt{2} \\ \Rightarrow \frac{\left(AB-AX\right)}{AB}=\frac{1}{\sqrt{2}} \\ \Rightarrow 1-\frac{AX}{AB}=\frac{1}{\sqrt{2}} \\ \Rightarrow \frac{AX}{AB}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{\left(2-\sqrt{2}\right)}{2} \end{gathered}$$