Question

AC and BD are chord of a circle which bisect each other prove that AC and BD are the diameters and ABCD is a rectangle

Answer 1

hey I am vipul here
Let AC and BD are the chords of a circle which bisect each other.

1. We have to prove that AC and BD are diameter

Since it is given that AC and BD are the chords of a circle which bisect each other.

Now in ΔABD,

∠A = 90

So BD is a diameter   (since angle in a semi-circle is 90)

Again in ΔBCD,

∠D = 90

So AC is a diameter   (since angle in a semi-circle is 90)

So AC and BD are the diameters.

2. We have to prove that ABCD is a rectangle

Let AC and BD are the chords of a circle which bisect each other at the point O.

Now in ΔOAB and ΔOCD

OA = OC   (given)

OB  = OD  (given)

∠AOB = ∠COD  (since vertically opposite angles)

So by SAS congruent criteria,

ΔOAB ≅ ΔOCD

BY CPCT

AB = CD  ...........1

Similarly we can show that

AD = CB ..........2

Add equation 1 and 2, we get

       AB + AD = CD + CB

=> ∠BAD = ∠BCD

So BD divides the circle into two equal semi-circle and the angle of each one is 90

So ∠A = 90 and ∠C = 90 ............3   

Similarly we can show that

∠B = 90 and ∠D = 90 .........4

From equation 3 and 4, we get

∠A = ∠B  =∠C = ∠D = 90

So ABCD is a rectangle.

Thanxx