Question

Suppose AB and CD are two diameters of a circle. Prove that the quadrilateral ACBD is a rectangle.

Answer 1

The easiest way to prove this is a rectangle is to draw it. I have attached a picture, but I will explain what the picture is showing.

When you create the edges that form the rectangle, you have a chord that is congruent to the other chord because the distances between AC and BD are the same. This is also true for the distances between AD and CB. The intersections of chords (other than the diameters) form a 90 degree angle.

The two angles that form each angle would need to add up to 90 degrees so the total number of degrees represented is 360 degrees.

Answer 2

Given : AB and CD are the diameters,

That is, AB = CD

We have to prove : ACBD is a rectangle,

Proof :

Since, CD is the diameter,

By the angle inscribed in semicircle theorem,

m∠DAC = m∠DBC = 90°

Similarly, AB is the diameter,

⇒ m∠ADB = m∠ACB = 90°

Now, In right triangle ADC,

`AD^2+AC^2=DC^2` ----(1) ( By Pythagoras theorem )

And, in right triangle ADB,

`AD^2+DB^2=AB^2`

`\implies AD^2+DB^2=DC^2` ----(2), ( Because, AB = CD )

By equation (1) - equation (2),

`AC^2-DB^2=0`

`AC^2=DB^2`

⇒ AC = BD

Similarly we can prove,

AD = BC,

Since, in the quadrilateral ACBD,

m∠DAC = m∠DBC = m∠ADB = m∠ACB = 90°,

AC = BD and AD = BC

⇒ ACBD is a rectangle.