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Home / Geometry / If two equal chords of a circle intersect each other, show that two segments of one are equal to two segments of the other.

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

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)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...