**Exercise-1: prove that, if two chords of a circle bisect each other , their point intersection will be the centre of the circle.**

**General enunciation**: We have to prove that, if two chords of a circle bisect each other, their point of intersection will be the centre of the circle.

**Particular enunciation**: Let in the circle ABCD the chords AB and CD bisect each other at E.

We have to prove that the point E is the centre of the circle ABCD.

**Construction**: Let the centre of the circle is O rather than E. We join O,E.

**Proof**: O be the centre of the circle. E be the midpoint of the chords AB and CD respectively.

∴OE⊥AB and OE⊥CD

The line segment joining the centre of a circle to midpoint of a chord other than diameter, is perpendicular to the chord.

So ∠OEC = 1 right angle

AB and CD are two intersector straight lines.

Both ∠OEA and ∠OEC are not equal to 1 right angle.

∴ No point other than E can be the centre of the circle.

∴E be the centre of the circle. [proved]

**Exercise-2: Prove that the line joining the midpoints of two parallel chords passes through the centre and is perpendicular to the two chords.**

**General enunciation**: we have to prove that the line joining the midpoints of two parallel chords passes through the centre and is perpendicular to the two chords.

**Particular enunciation**: Let in the circle of centre O, AB and CD are two parallel chords. E and F are the midpoints of AB and CD respectively. The line segment EF is the joining line of the midpoints E and F. It is required to prove that EF passes through the centre and is perpendicular to the chords.

**Construction**: We join O and F.

**Proof**: O be the centre of the circle and E and F be the midpoints of the chords AB and CD.

∴OE⊥AB and OF⊥CD

The line segment joining the centre of the circle to midpoint of a chord other than diameter, is perpendicular to the chord.

So ∠OEB = 1 right angle

And ∠OFD= 1 right angle

∴∠OEB=∠OFD.

The two angles ∠OEB and ∠OFD are corresponding and equal.

∴OF and OE will be the transversals [AB∥CD]

i.e. lie on the same straight line [O be the centre of the circle]

therefore EF passes through the centre (proved)

**Exercise -3: Two chords AB and AC of a circle make circle make equal angles with the radius through A. Prove that AB=AC.**

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

**Particular enunciation**: Let O is the centre of the circle ABC. The two chords AB and AC has made equal angles ∠BAO and ∠CAO with the radius OA through A.

It is required to prove that AB=AC.

**Construction**: from the centre O we draw the perpendicular OM and ON respectively to AB and AC.

**Proof**: O is the centre and OM⊥AB

The perpendicular from the centre of a circle to a chord bisects the chord.

∴AM= ½ AB Again ON⊥AC

∴AN= ½ AC

In ∆AOM and ∆AON

∠AMO=∠ANO [1 right angle]

∠MAO=∠NAO [supposition]

And AO=AO [common side]

So, ∆AOM≌∆AON [ASA theorem]

∴AM=AN

Or, ½ AB= ½ AC [from 1]

Or, AB=AC [proved]

**Exercise-4: In the figure , O is the centre of circle and the chord AB = chord AC . Prove that ∠BAO=∠CAO.**

**General Enunciation**: In the figure , O is the centre of circle and the chord AB = chord AC . We have to prove that ∠BAO=∠CAO.

**Particular enunciation**: Given that O is the centre of the circle and the chord AB=AC. We join O and A.

We have to prove that ∠BAO=∠CAO

**Construction**: We join O, B and O, C

**Proof**: In ∆AOB and ∆AOC

AB=AC [given]

OB=OC [radius of same circle]

OA=OA [common side]

So ∆AOB≌∆AOC [SSS theorem]

∴∠BAO=∠CAO. [proved]

**Exercise-5: A circle passes through the vertices of a right – angled triangle. Show that the centre lies on the midpoint of the hypotenuse.**

**General enunciation**: A circle passes through the vertices of a right-angled triangle. It is required to show the centre lies on the midpoint of the hypotenuse.

**Particular enunciation**: Let ABC is a right – angled triangle. Its ∠ABC= 1 right angle and AC is the hypotenuse, The circle passes through the vertices A,B and C.

**Construction**: From O we draw the perpendicular OD to AB and the perpendicular OE to BC. Again we join O, B.

**Proof**: O is the centre of the circle and OD⊥AB

The perpendicular from the centre of a circle to a chord bisects the chord]

∴AD=BD ——————————–(1)

In ∆AOD and ∆BOD

AD=BD [from 1]

OD=OD [common side]

And ∠ADO=∠BDO= 1 right angle [OD⊥AB]

So ∆AOD≌∆BOD [SAS theorem]

∴AO=BO

Similarly from ∆BOE and ∆COE it is proved that CO=BO

So AO=BO=CO ——————————-(2)

Therefore O is the centre of the circle.

Hypotenuse AC=AO+CO

=CO+CO [from 2]

=2CO

Or, 2CO=AC

Or, CO= ½ AC

Therefore O lies on the midpoint of the hypotenuse AC. (proved)

**Exercise-6: A chord AB of one of two concentric circles intersect the other at C and D. Prove that AC=BD.**

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

**Particular enunciation**: Let O be the two circles ABE and CDF. The chord AB of the circle ABE intersects the circle CDF at C and D.

It is required to prove that AC=BD.

**Construction**: From O we draw the perpendicular OP on AB or CD.

**Proof**: O is the centre and OP⊥CD, OP⊥AB.

The perpendicular from the centre of a circle to a chord bisects the chord.

∴CP=PD and AP=BP ————————- (1)

Again, AP=AC+CP

And BP=PD+BD

Since AP=BP [from 1]

So AC+CP=PD+BD

Or, AC+CP=CP+BD [CP=PD]

Or, AC+CP=BD+CP

Or, AC=BD. (proved)