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Problem 188

# Given: Point $$P$$ is in the interior of $$\triangle A B C$$. Prove: $\mathrm{AP}+\mathrm{PB}+\mathrm{PC}>1 / 2(\mathrm{AB}+\mathrm{AC}+\mathrm{BC})$

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We are given that point P is inside $$\triangle ABC$$. To prove $$\mathrm{AP}+\mathrm{PB}+\mathrm{PC} > \frac{1}{2}(\mathrm{AB}+\mathrm{AC}+\mathrm{BC})$$, apply the triangle inequality theorem to the smaller triangles $$\triangle APB$$, $$\triangle APC$$, and $$\triangle BPC$$. Add the resulting inequalities, simplify, and divide by 2 to reach the desired result.
See the step by step solution

## Step 1: 1. Write down given information and requirements

We are given that point P is inside the triangle ABC. And we need to prove: $$\mathrm{AP}+\mathrm{PB}+\mathrm{PC} > \frac{1}{2}(\mathrm{AB}+\mathrm{AC}+\mathrm{BC})$$

## Step 2: 2. Apply triangle inequality theorem to triangles containing point P

We can see three smaller triangles: $$\triangle APB$$, $$\triangle APC$$, and $$\triangle BPC$$. For each of these triangles, we can apply the triangle inequality theorem: a) For $$\triangle APB$$: $\mathrm{AP} + \mathrm{PB} > \mathrm{AB}$ b) For $$\triangle APC$$: $\mathrm{AP} + \mathrm{PC} > \mathrm{AC}$ c) For $$\triangle BPC$$: $\mathrm{PB} + \mathrm{PC} > \mathrm{BC}$

## Step 3: 3. Add inequalities (a), (b), and (c)

Now, let's sum the inequalities from step 2: $$(\mathrm{AP} + \mathrm{PB} > \mathrm{AB}) + (\mathrm{AP} + \mathrm{PC} > \mathrm{AC}) + (\mathrm{PB} + \mathrm{PC} > \mathrm{BC})$$

## Step 4: 4. Simplify the inequality

Combine the terms on both sides of the inequality: $$2(\mathrm{AP} + \mathrm{PB} + \mathrm{PC}) > (\mathrm{AB} + \mathrm{AC} + \mathrm{BC})$$

## Step 5: 5. Divide both sides by 2

Finally, divide both sides of the inequality by 2 to reach the desired result: $$\mathrm{AP} + \mathrm{PB} + \mathrm{PC} > \frac{1}{2}(\mathrm{AB} + \mathrm{AC} + \mathrm{BC})$$ Thus, we have proven that the sum of the lengths of the segments AP, PB, and PC is greater than half the perimeter of triangle ABC.

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