**Problem: A chord AB one of the two concentric circles intersect the other circle at points C and D. Prove that AC = BD.**

**General enunciation**: A chord AB one of the two concentric circles intersect the other circle at points C and D. We have to prove that AC = BD.

**Particular enunciation**: Let O is the common centre of the circles EAB and FCD. The chords AB of EAB intersect the circle FCB at the points C and D. Let us prove that AC = BD.

**Construction**: Let us draw OP perpendicular to AB or CD.

**Proof**: Since O is the centre the circle and OP ⊥ CD. Then we get CP = PD.

Similarly, O is the centre of the circle and OP ⊥ AB. Then we get AP = PB.

But, AP = AC + CP and PB = PD + BD.

∴ AC + CP = PD + BD

⟹ AC + CP = CP + BD [∵ CP = PD]

∴ AC = BD. **(Proved)**

**Problem : If two equal chords of a circle intersect each other, show that two segments of one are equal to two segments of the other.**

General enunciation: If two equal chords of a circle intersect each other, show that two segments of one are equal to two segments of the other.

**Particular enunciation**: Suppose O is the centre of the circle MQNP. Two equal chords MN and PQ intersect each other at the point A. Let us prove that MA = PA and AQ = AN.

**Construction**: Let us draw two perpendiculars OC and OB to the chords MN and PQ respectively. (O, A) are joined.

**Proof**: Since equal chords are equidistant from the centre.

So, in right angle **∆ **COA and **∆ **BOA we get

OC = OB and OA is the common hypotenuse.

∴ **∆ **COA ≅ **∆ **BOA

∴ AC = AB —————————(i)

Now, since O is the centre of the circle and OC ⊥ MN then we get MC = ½ MN.

Similarly, O is the centre of the circle and OB ⊥ PQ then we get PB = ½ PQ.

But given that MN = PQ ——————(ii)

∴ MC = PB ————————————(iii)

Adding equations (i) and (iii) we get

AC + MC = AB + PB

⟹ AM = AP ———————————-(iv)

Again, subtracting (iv) from (ii) we get

MN – AM = PQ – AP

⟹ AN = AQ

∴ AM = AP and AN = AQ. **(Proved)**

**Problem: Two chords AB and AC of a circle subtend equal angles with the radius passing through A. Prove that AB = AC.**

**General enunciation**: Two chords AB and AC of a circle subtend equal angles with the radius passing through A. We have to prove that AB = AC.

**Particular enunciation**: Suppose O is the centre of the circle ABC. Two chords AB and AC of the circle subtend equal angles with the radius OA, i.e. ∠OAB = ∠OAC. Let us prove that AB = AC.

**Construction**: Let us draw two perpendiculars OM and ON to the chords AB and AC respectively.

**Proof**: Since O is the centre of the circle ABC and OM ⊥ AB. Then we get

AM = ½ AB.

Similarly, O is the centre of the circle ABC and ON ⊥ CD. Then we get

AN = ½ AC.

Now in **∆ **OAM and** ∆** OAN we get ∠OAM = ∠OAN [∵∠OAB = ∠OAC]

and ∠OMA = ∠ONA [∵each of them is one right angles]

and OA is common perpendicular.

∴ **∆**OAM≅**∆**OAN

Then we get AM = AN

⟹ ½ AB = ½ AC

∴ AB = AC. **(Proved)**