Mathematics
If P is a point on a circle with centre O. If P is equidistant from the two radii OA and OB, prove that arc AP = arc PB.
Answer
In △ OMP and △ ONP,
⇒ ∠OMP = ∠ONP = 90°
⇒ OP = OP (Hypotenuse)
⇒ PM = PN (Given)
∴ △ OMP ≅ △ ONP
Since, the triangles are congruent, their corresponding parts are equal.
∠MOP = ∠NOP
∠AOP = ∠BOP
Equal angles at the centre of a circle subtend equal arcs.
∴ arc AP = arc PB.
Hence, proved that arc AP = arc PB.
Related Questions
In the given figure, P is the mid-point of arc APB and M is the mid-point of chord AB of a circle with centre O. Prove that:
(i) PM ⟂ AB
(ii) PM produced will pass through the centre O
(iii) PM produced will bisect the major arc AB.
Prove that in a cyclic trapezium, the non-parallel sides are equal.
In the given figure, two chords AC and BD of a circle intersect at E. If arc AB = CD, prove that : BE = EC and AE = ED.
In the given figure, two chords AB and CD of a circle intersect at a point P.
If AB = CD, prove that: arc AD = arc CB.
