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$P$ is any point in the interior of triangle $ABC$. Prove that $AB + AC > BP + CP$.

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Answer:

Step by Step Explanation:

1. Let us first mark the point $P$ in the interior of $\triangle ABC$.
2. Now, let us join line $BP$ and $CP$ and produce $CP$ to meet $AB$ at $Q$.
3. 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 $$\begin{aligned} & BQ + PQ > BP && \ldots (2) \end{aligned}$$
4. By adding (1) and (2), we get: $$\begin{aligned} & AC + AQ + BQ + PQ > CP + PQ + BP \\ \implies & AC + AQ + BQ > CP + BP && [\text{Subtrating PQ from both the sides}] \\ \implies & AC + AB > CP + BP && [\text{As AQ + BQ = AB}] \end{aligned}$$
5. Thus, we have $$\bf {AC + AB > CP + BP}$$ .

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