Chapter 15.2, Problem 21PSC

To determine

To prove: that the function $PY+PZ<XY+XZ$ .

Answer to Problem 21PSC

  $PY+PZ<XY+XZ$ is proved.

Explanation of Solution

Given:

A point P in the interior of the $\Delta XYZ$ .

Calculation:

A point P in the interior of the $\Delta XYZ$ .

  

Now, it is to prove $PY+PZ<XY+XZ$ .

As the sum of two sides of any triangle is greater than its third side. Extend $\overline{PY}\text{to Q on }\overline{XZ}$ . Then,

  $\begin{array}{l}XY+XQ>YP+PQ\{PQ+QZ>PZ\}\\ XY+XQ>QZ+PQ\{PY+PQ>QZ\}\\ XY+\left(XQ+QZ\right)>YZ+PZ\\ \therefore XY+XZ>YZ+PZ\end{array}$

Hence, the function $PY+PZ<XY+XZ$ is proved.