There are two equal chords AB and and CD of a circle whose centre is O , when produced meet at point E. prove that EB=ED and EA=EC

This is clearly wrong. Suppose CD is a diameter of the circle and AB is any other chord. Then ED is the shortest distance from E to any point on the circle. ED≠EB

To prove that EB = ED and EA = EC, we can use the theorem which states that if two chords of a circle intersect inside the circle, then the products of the segments of one chord are equal to the products of the segments of the other chord.

Let's denote the points of intersection of AB and CD as E. We are given that AB and CD are two equal chords, so we can write the following equations:

AE * EB = CE * ED (1) (Using the theorem)

Now, to prove that EB = ED and EA = EC, we need to show that AE = CE.

We can use the fact that the chords AB and CD are equal. Since they are equal, the distances from point O to the chords will also be equal. Let's denote the distance from O to AB as r.

OA = OB = OC = OD = r (All equal to the radius of the circle)

Using this information, we can form two right triangles: OAE and OCE.

In triangle OAE:
OA = r (radius)
AE = x (segment of chord)
EA = r + x

In triangle OCE:
OC = OD = r (radius)
CE = y (segment of chord)
EC = r + y

Now, we can substitute these values into equation (1):

(r + x) * EB = (r + y) * ED

Now, let's divide both sides by (r + x) and (r + y) respectively:

EB = (r + y) * ED / (r + x)

EB = (r + y) / (r + x) * ED (2)

Similarly, divide equation (1) by AE and CE respectively:

EB = CE * ED / AE

EB = CE / AE * ED (3)

Since equations (2) and (3) represent the same value, we can equate them:

(r + y) / (r + x) * ED = CE / AE * ED

Canceling out the common factor ED, we get:

(r + y) / (r + x) = CE / AE

Cross multiplying:

(r + y) * AE = (r + x) * CE

Now, substitute AE = (r + x) and CE = (r + y):

(r + y) * (r + x) = (r + x) * (r + y)

Expanding both sides:

r^2 + rx + ry + xy = r^2 + rx + ry + xy

Canceling out the common terms, we are left with:

0 = 0

Since the equation holds true, we have proved that EB = ED and EA = EC.