Question

Theorem 10.1 Equal chords of a circle subtend equal angles at the centre?
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Answer 1

Proof: Given, in ∆AOB and ∆POQ,

AB = PQ (Equal Chords) …………..(1)

OA = OB= OP=OQ (Radii of the circle) ……..(2)

From eq 1 and 2, we get;

∆AOB ≅ ∆POQ (SSS Axiom of congruency)

Therefore, by CPCT (corresponding parts of congruent triangles), we get;

∠AOB = ∠POQ

Hence, Proved.

“If two angles subtended at the center by two chords are equal, then the chords are of equal length.”

Proof: Given, in ∆AOB and ∆POQ,

∠AOB = ∠POQ (Equal angle subtended at centre O) ……………(1)

OA = OB = OP = OQ (Radii of the same circle) ……………(2)

From eq. 1 and 2, we get;

∆AOB ≅ ∆POQ (SAS Axiom of congruency)

Hence,

AB = PQ (By CPCT)

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

circle theorem 2

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)