Question

Show that the difference of any two sides of a triangle is less than the third side.

Answer 1

$$\begin{gathered} \mathrm{Construction}:\mathrm{Let}\mathrm{AC}\hspace{0.17em}>\mathrm{AB}. \\ \mathrm{Then}\mathrm{along}\mathrm{AC},\mathrm{set}\mathrm{off}\mathrm{AD}=\mathrm{AB}.\mathrm{Join}\mathrm{BD}. \\ \mathrm{Proof}\hspace{0.17em}:\mathrm{AB}=\hspace{0.17em}\mathrm{AD}\Rightarrow \angle 1=\angle 2 \\ \mathrm{Side}\mathrm{CD}\mathrm{of}∆\mathrm{BCD}\mathrm{has}\mathrm{been}\mathrm{produced}\mathrm{to}\mathrm{A}. \\ \therefore \angle 2>\angle 4[\because \mathrm{ext}\mathrm{angle}>\mathrm{each}\mathrm{int}.\mathrm{opposite}\mathrm{angle}] \\ \mathrm{Again},\mathrm{side}\mathrm{AD}\mathrm{of}△\mathrm{ABD}\mathrm{has}\mathrm{been}\mathrm{produced}\mathrm{to}\mathrm{C}. \\ \therefore \angle 3>\angle 1[\because \mathrm{ext}\mathrm{angle}>\mathrm{each}\mathrm{int}.\mathrm{opposite}\mathrm{angle}] \\ \mathrm{Consequently},\angle 3>\angle 2[\because \angle 1=\angle 2] \\ \mathrm{Now},\angle 3>\angle 2\mathrm{and}\angle 2>\angle 4\Rightarrow \angle 3>\angle 4 \\ \therefore \mathrm{BC}>\mathrm{CD}[\mathrm{side}\mathrm{opposite}\mathrm{to}\mathrm{the}\mathrm{greater}\mathrm{angle}\mathrm{is}\mathrm{longer}] \\ \Rightarrow \mathrm{CD}<\mathrm{BC} \\ \Rightarrow \mathrm{AC}-\mathrm{AD}<\mathrm{BC} \\ \Rightarrow \mathrm{AC}-\mathrm{AB}<\mathrm{BC}[\because \mathrm{AD}=\mathrm{AB}] \\ \mathrm{Hence},\mathrm{AC}-\mathrm{AB}<\mathrm{BC} \\ \mathrm{Similarly},\mathrm{BC}-\mathrm{AC}<\mathrm{AB}\mathrm{and}\mathrm{BC}-\mathrm{AB}<\mathrm{AC}. \end{gathered}$$