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Home / Geometry / Circle| Part – 2

# Circle| Part – 2

Exercise : Two chords AB and AC of  a circle make circle make equal angles with the radius through A. 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 —————————– (1)

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: In the figure , O is the centre of circle and the chord AB = chord AC . 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]

OA=OA   [common side]

So ∆AOB≌∆AOC [SSS theorem]

∴∠BAO=∠CAO.  [proved]

Ex: 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]

In ∆AOD and ∆BOD

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: 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 intersects 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)

## 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 ...