State true or false.If two equal chords of a circle intersect, then the parts of one chord are separately equal to the parts of the other chord.

Given that, two equal chords of a circle intersect.
Assuming that

A B

and

C D

are two equal chords meeting at point E of a circle with center O.
Drawing

O M ⊥ A B a n d O N ⊥ C D,

and joining

O E .

Considering triangle

O M E

and

O N E,

As we know when chords equidistant from the center of a circle are equal in length, then

O M = O N

O E = O E

(Common side)
And

∠ O M E = ∠ O N E = 90 ° .

By RHS congruence rule,

Δ O M E ≅ Δ O N E

E M = E N ...... 1

(C.P.C.T)

A B = C D

(C.P.C.T)
As we know perpendicular drawn at the center of the circle to a chord bisects the chord, so we have:

A M = M B = A B 2

and,

C N = N D = C D 2

Therefore,

A M = C N ...... 2

Adding equation 1 and equation 2.

E M + A M = E N + C N

A E = C E

Subtracting both sides by

A E from A B = C D .

A B − A E = C D − A E

Using the figure we have,

A E = C E

C D − C E = D E

A B − A E = B E

So,

A B − A E = C D − A E

B E = C D − C E

B E = D E

Since

B E = D E a n d A E = C E,

thus the statement is true.
Hence, option 1 is correct.

State true or false.

If two equal chords of a circle intersect, then the parts of one chord are separately equal to the parts of the other chord.