Vanuatu
^@P^@ is any point in the interior of triangle ^@ABC^@. Prove that ^@AB + AC > BP + CP^@.
Answer:
- Let us first mark the point ^@P^@ in the interior of ^@\triangle ABC^@.
- Now, let us join line ^@BP^@ and ^@CP^@ and produce ^@CP^@ to meet ^@AB^@ at ^@Q^@.
- We know that the sum of two sides of a triangle is greater than the third side.
Thus in ^@ \triangle ACQ,^@ we have @^ \begin{aligned} & AC + AQ > CQ \\ \implies & AC + AQ > CP + PQ && \ldots (1) \end{aligned} @^ Similarly in ^@ \triangle BPQ,^@ we have @^ \begin{aligned} & BQ + PQ > BP && \ldots (2) \end{aligned} @^ - By adding (1) and (2), we get:@^\begin{aligned} & AC + AQ + BQ + PQ > CP + PQ + BP \\ \implies & AC + AQ + BQ > CP + BP && [\text{Subtrating PQ from both the sides}] \\ \implies & AC + AB > CP + BP && [\text{As AQ + BQ = AB}] \end{aligned}@^
- Thus, we have @^ \bf {AC + AB > CP + BP} @^.
