Mathematics
Prove that a diameter of a circle, which bisects a chord of the circle, also bisects the angle subtended by the chord at the centre of the circle.
Answer
Let POQ be a diameter, bisecting chord AB at L.
Join OA and OB
In △OLA and △OLB:
OA = OB [Both are radii of the same circle]
AL = LB [PQ bisects the chord AB]
OL = OL [common side]
△OLA ≅ △OLB [By the SSS rule]
Since the triangles are congruent, their corresponding parts must be equal:
∠AOL = ∠BOL
This proves that the diameter PQ bisects ∠AOB, which is the angle subtended by the chord AB at the center O.
Hence, proved that the diameter also bisects the angle subtended by the chord at the centre.
Related Questions
AB and CD are two parallel chords of a circle and a line l is the perpendicular bisector of AB. Show that l is the perpendicular bisector of CD also.
Prove that diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.
In the given figure, L and M are mid-points of two equal chords AB and CD of a circle with centre O. Prove that :
(i) ∠OLM = ∠OML
(ii) ∠ALM = ∠CML
In the given figure, AB and AC are equal chords of a circle with centre O and OP ⟂ AB, OQ ⟂ AC. Prove that PB = QC.
