Question

Show that if two chords of a circle bisect one other they must be the diameters of the circle.

Answer 1

Let AB and CD be two chords intersecting at point O. Join AC and BD.

Now ΔAOC≈ΔBOD

⇒AC=BD

⇒∧AC =∧ BD-----------(1)

Now,

 ΔAOD≈ΔBOC

⇒AD=BC

⇒∧AD=∧BC--------(2)

(1)+(2)

⇔∧AC+∧AD =∧BD+∧BC

⇒ ANGLE "CAD"= ANGLE "CBD"

Then CD divides the circle into two equal parts thus CD is a diameter

Similarly AB is also the diameter and they both meet at point O

Thus they bisect each other....

Refer the attachment for the diagram

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Answer 2

Hello Mate!

Given : Chord AC bisect Chord BD.

To prove : AC and BD are diameters.

To construct : Join AB, BC, CD and AD to be more precise.

Proof : In ∆AOB and ∆COD we have,

AO = OC [ Given ]

< AOB = < COD [ Vertically opp. angles ]

OD = OB [ Given ]

Hence, ∆AOB ~ ∆COD by SAS congruency.

AB = CD [ c.p.c.t ]

arc(AB) = arc(CD) _(i)

In ∆AOD and ∆BOC we have,

OD = OB [ Given ]

< AOD = < BOC [ Vertically opp. angles ]

OA = OC [ Given ]

Hence, ∆AOD ~ ∆BOC by SAS congruency.

AD = BC [ c.p.c.t ]

arc(AD) = arc(BC) _(ii)

By adding (i) and (ii) we get,


arc(AB) + arc(AD) = arc(CD) + arc(BC)

arc(BAD) = arc(BCD)

BD = AC

Q.E.D

Have great future ahead!