Suppose ab and cd are two diameters of a circle. prove that the quadrilateral acbd is a rectangle.

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,

----(1)  ( By Pythagoras theorem )

And, in right triangle ADB,

----(2),  ( Because, AB = CD )

By equation (1) - equation (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.

1 is c

2 is a

3 is a

step-by-step explanation: