^@P^@ is any point in the interior of triangle ^@ABC^@. Prove th | Math Question and Answer | Edugain Bahrain

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

Answer:

Step by Step Explanation:

Let us first mark the point $P$ $P$ in the interior of $△ A B C$ $△ A B C$ .
Now, let us join line $B P$ $B P$ and $C P$ $C P$ and produce $C P$ $C P$ to meet $A B$ $A B$ at $Q$ $Q$ .
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 $\bf {AC + AB > CP + BP}$ .

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