Question 692844

first draw the picture:

{{{drawing( 600,600, -5, 15, -5, 10, 
         grid(0),blue( circle( 5.5, 2.5, 4.9 ) ),locate( 5.5, 2.5, O ),locate( 1, 1, A ),locate(10, 1, B),locate( 10, 4, C ),locate( 1, 4, D ),
         line( 1, 1, 10, 1 ),line( 10, 1, 1, 4 ),line( 1, 1, 10, 4 ),line( 10, 1, 10, 4 ),line(1, 1, 1, 4 ),line( 1, 4, 10, 4 )
)}}}


next, we will need to prove this:
theorem: Diagonal of any rectangle inscribed in a circle is a diameter of the circle. (

This is essentially the converse of Thales' theorem.

Let A, B, C and D be the vertices of a rectangle inscribed in a circle and let AC and BD be the diagonals of this rectangle. 

We can now focus on any one of the four triangles: 

{{{ABC}}}, {{{BCD}}}, {{{CDA}}} and {{{DAB}}}

Let's take {{{ABC}}}:

{{{AC}}} side of this triangle is the {{{diagonal}}} of the rectangle

{{{AB}}} and {{{BC}}} are two sides of the rectangle, which potentially may be of different length (but we will prove they must be the {{{same}}} if the area of the inscribed rectangle is maximized).
The angle at {{{B}}} is the right angle since it is one of the angles of the rectangle, which by definition has four right angles.

Hence, looking at the triangle{{{ ABC}}}, we can see that this is a {{{right-angled}}} triangle inscribed in the circle.

Therefore to prove (1), we need to show that side{{{ AC}}} of any such triangle must be a  diameter of circle.

Proof: 

Choose any three points {{{A}}}, {{{B}}} and {{{C}}} on the circle and connect these points to make a triangle {{{ABC}}}.

Let's suppose that the claim is that the angle at {{{B}}} is right angle.

We will show that if this is true then it must follow that {{{AC}}} is a {{{diameter}}} of the {{{circle}}}.
Connect the center of the circle ({{{O}}}) with each of the vertices of the triangle creating the segments {{{OA}}}, {{{OB}}} and {{{OC}}}.

Let's call the angle defined by the path {{{OAB}}} as {{{alpha}}} and the angle defined by the path {{{OCB}}} as {{{beta}}}.

Since {{{OA}}}, {{{OC }}}and{{{ OB}}} are all of {{{equal}}} length (equal to the length of the  radius  of the circle ) then {{{all}}}{{{ three}}}{{{ inner}}}{{{ triangles}}} ({{{OAB}}}, {{{OAC}}} and {{{OBC}}}) are {{{isosceles}}}.

We will next want to find the angle between {{{OA}}} and {{{OC }}}(the angle at {{{O}}} made out by segments {{{OA}}} and {{{OC}}}) using only angles {{{alpha}}} and {{{beta}}}as given.
The angle between {{{OA}}} and {{{OB}}} is equal to {{{180 - 2alpha}}}° (due to the fact that {{{OAB}}} is {{{isosceles}}} and the fact that the sum of all three internal angles in a triangle sum to {{{180}}}°).

Similarly, the angle between {{{OB}}} and {{{OC}}} is equal to  {{{180 - 2beta}}}°.

Finally, since all the three angles at {{{O}}} add up to {{{360}}}° (full circle), it follows that the angle between {{{OA}}} and {{{OC}}} is equal to {{{360 - (180 - alpha) – (180 - beta) = 2(alpha+beta)}}}.

However, the angle at {{{B}}} of the original triangle {{{ABC}}} is equal to {{{(alpha+beta)}}} (follows from the fact that {{{OAB}}} and {{{OBC}}} triangles are {{{isosceles}}}).

This angle was claimed to be {{{90}}}° and therefore the angle between {{{OA}}} and {{{OC}}} is equal to {{{180}}}°.

This means that the points {{{A}}} and {{{C}}} and the {{{center}}} of the circle ({{{O}}}) are {{{collinear}}}.

In other words, the {{{center}}} of the circle is {{{lying}}} on the straight line segment {{{AC}}}, which in turn means that {{{AC}}}{{{ must}}} be a {{{diameter }}}of the{{{ circle}}}.

So, we can conclude that {{{each}}}{{{ diagonal}}} of the rectangle inscribed in a circle {{{must}}} be a circles {{{diameter}}}.