Question

PROVE THAT “The perpendicular to a chord bisects the chord if drawn from the centre of the circle.”
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Answer 1

Answer:

ANSWER To prove that the perpendicular from the centre to a chord bisect the chord.

Consider a circle with centre at O and AB is a chord such that OX perpendicular to AB

To prove that   AX=BX

In ΔOAX and ΔOBX

∠OXA=∠OXB  [both are 90 ]

OA=OB  (Both  are radius of circle )

OX=OX  (common side )

ΔOAX≅ΔOBX

AX=BX  (by property of congruent triangles )

hence proved.

Step-by-step explanation:

I hope this will help you .........

Answer 2

Step-by-step explanation:

In the above figure, as per the theorem, OD ⊥ AB, therefore, AD = DB.

Proof: Given, in ∆AOD and ∆BOD,

∠ADO = ∠BDO = 90° (OD ⊥ AB) ………(1)

OA = OB (Radii of the circle) ……….(2)

OD = OD (Common side) ………….(3)

From eq. (1), (2) and (3), we get;

∆AOB ≅ ∆POQ (R.H.S Axiom of congruency)

Hence, AD = DB (By CPCT)