SOLUTION: A pleasant day Sir/ Madam, my daughter has this problem in geometry and I am not good in geometric proofs, can you please help me? the illustration or drawing is a rectangle ins

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Question 692844: A pleasant day Sir/ Madam, my daughter has this problem in geometry and I am not good in geometric proofs, can you please help me?
the illustration or drawing is a rectangle inscribed in a circle
Given: Quadrilateral ABCD is a rectangle
Prove: line segment AC and line segment BD are diameters of the circle
Thank you very much for anyone who can help me!

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

first draw the picture:


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 betaas 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+-+%28180+-+alpha%29+%96+%28180+-+beta%29+=+2%28alpha%2Bbeta%29.
However, the angle at B of the original triangle ABC is equal to %28alpha%2Bbeta%29 (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.