Question

In $∆$ABC, sides AB and BC are congruent
(i) A-P-C, then show that BP < congruent sides.
(ii) If A-C-P, then show that BP > congruent sides.

Answer 1

$$\begin{gathered} \mathrm{AB}=\mathrm{BC} \\ \left(\mathrm{i}\right)\mathrm{Given}:\mathrm{A}-\mathrm{P}-\mathrm{C} \end{gathered}$$

$$\begin{gathered} \mathrm{Let}\angle \mathrm{BAC}=\angle \mathrm{BCA}=\mathrm{\theta } \\ \therefore \angle \mathrm{ABC}=180°-(\angle \mathrm{BAC}+\angle \mathrm{BCA}) \\ \Rightarrow \angle \mathrm{ABC}=180°-(\mathrm{\theta }+\mathrm{\theta }) \\ \Rightarrow \angle \mathrm{ABC}=180°-2\mathrm{\theta } \\ \mathrm{Now},\mathrm{in}△\mathrm{ABC}, \\ \angle \mathrm{ABP}<\angle \mathrm{ABC} \\ \Rightarrow \angle \mathrm{ABP}<180°-2\mathrm{\theta }......\left(\mathrm{i}\right) \\ \mathrm{Also},\mathrm{in}△\mathrm{ABP}, \\ \angle \mathrm{APB}=180°-(\angle \mathrm{PAB}+\angle \mathrm{ABP}) \\ \Rightarrow \angle \mathrm{APB}=180°-\mathrm{\theta }-\angle \mathrm{ABP} \\ \Rightarrow \angle \mathrm{APB}>180-\mathrm{\theta }-(180-2\mathrm{\theta }) \\ \therefore \angle \mathrm{APB}>\mathrm{\theta } \\ \mathrm{Thus},\angle \mathrm{APB}>\angle \mathrm{PAB} \\ \Rightarrow \mathrm{AB}>\mathrm{PB} \end{gathered}$$

(ii)


$$\begin{gathered} \mathrm{In}△\mathrm{ABC}, \\ \mathrm{let}\angle \mathrm{BAC}=\angle \mathrm{BCA}=\mathrm{\theta } \\ \mathrm{We}\mathrm{have} \\ \mathrm{\theta }+\mathrm{\theta }+\angle \mathrm{ABC}=180° \\ \Rightarrow 2\mathrm{\theta }+\angle \mathrm{ABC}=180° \\ \Rightarrow 2\mathrm{\theta }<180° \\ \Rightarrow \mathrm{\theta }<90° \\ \Rightarrow -\mathrm{\theta }>-90° \\ \Rightarrow 180°-\mathrm{\theta }>90° \\ \Rightarrow \angle \mathrm{BCP}>90° \\ \mathrm{Therefore},△\mathrm{BCP}\mathrm{is}\mathrm{an}\mathrm{obtuse}\mathrm{triangle}. \\ \mathrm{The}\mathrm{side}\mathrm{opposite}\mathrm{the}\mathrm{greatest}\mathrm{angle}\mathrm{is}\mathrm{largest}.\text{T}\mathrm{hus},\mathrm{BP}\mathrm{is}\mathrm{the}\mathrm{largest}\mathrm{side}. \\ \mathrm{Thus},\mathrm{BP}>\mathrm{BC}. \end{gathered}$$