Chapter 15, Problem 25RP

To determine

To prove: that the sum of the distance from P to the four vertices $\left(PW+PX+PY+PZ\right)$ is greater than or equal to the sum of the diagonals.

Answer to Problem 25RP

  $PW+PY+PX+PZ\ge XZ+WY$ .

Explanation of Solution

Given:

P is any points inside the quadrilateral $WXYZ$ .

Calculation:

According to the question, P is any points inside the quadrilateral $WXYZ$ . There are three points P with the quadrilateral.

Case 1: P is not on either diagonal.

Now, the pictorial representation for this case is:

  

  $\begin{array}{l}\text{In}\Delta PXZ,PX+PZ>XZ\\ \text{In}\Delta PWY,PW+PY>WY\\ \therefore PX+PZ+PW+PY>XZ+WY\end{array}$

Case 2: P is on one diagonal.

Now, the pictorial representation for this case is:

  

  $\begin{array}{l}\text{In}\Delta PWY,PW+PY>WY\\ \text{In}\Delta PWY,PX+PZ>XZ\\ \therefore PW+PY+PX+PZ>XZ+WY\end{array}$

Case 3: P is on both diagonals.

Now, the pictorial representation for this case is:

  

  $\begin{array}{l}PW+PY=WY\\ PX+PZ=XZ\\ \therefore PW+PY+PX+PZ\ge XZ+WY\end{array}$

Thus, $PW+PY+PX+PZ\ge XZ+WY$

Hence, from both the case, it is proved that four vertices $\left(PW+PX+PY+PZ\right)$ is greater than or equal to the sum of the diagonals.