Chapter 9.4, Problem 36PSD

a.

To determine

To show:The squares of sides of two sides is equal to sum of squares of other two sides when point P is interior of rectangle ABCD .

Explanation of Solution

Given information:

P is any point in the interior of rectangle ABCD .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Proof:

  

  $\begin{array}{l}B{P}^2=y^2+T{B}^2\\ P{D}^2=x^2+D{S}^2\\ B{P}^2+P{D}^2=x^2+y^2+D{S}^2+T{B}^2\\ A{P}^2=y^2+A{T}^2\\ C{P}^2=x^2+S{C}^2\\ A{P}^2+C{P}^2=x^2+y^2+S{C}^2+A{T}^2\end{array}$

If $\overline{ST}$ is drawn through P , parallel to CB , STBC is a rectangle,

  $\overline{TB}\cong \overline{SC}$ and $\overline{DS}\cong \overline{AT}$ .

The sum of squares of BP and PD is equal to sum of squares of AP and CP .

b.

To determine

To show: The squares of sides of two sides is equal to sum of squares of other two sides when point P is exterior of rectangle ABCD .

Explanation of Solution

Given information:

P is any point exterior of rectangle ABCD .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

Proof:In the figure, point P is exterior of rectangle ABCD .

  

  $\begin{array}{l}B{P}^2=y^2+T{B}^2\\ P{D}^2=x^2+D{S}^2\\ B{P}^2+P{D}^2=x^2+y^2+D{S}^2+T{B}^2\\ A{P}^2=y^2+A{T}^2\\ C{P}^2=x^2+S{C}^2\\ A{P}^2+C{P}^2=x^2+y^2+S{C}^2+A{T}^2\end{array}$

If $\overline{ST}$ is drawn through P , parallel to CB , STBC is rectangle,

  $\overline{TB}\cong \overline{SC}$ and $\overline{DS}\cong \overline{AT}$ .

Yes, the result is same when point P is exterior of rectangle ABCD.