Chapter 11: Chapter 11

Problem 188

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

Short Answer

Expert verified

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.

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Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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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Chapter 11: Chapter 11

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

Short Answer

Step by step solution

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Step 1: 1. Write down given information and requirements

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

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

Step 4: 4. Simplify the inequality

Step 5: 5. Divide both sides by 2

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