Question

Equal chords AB and CD of a circle with centre 0, intersect at P
inside the circle. Show that OP bisects 2CPB.​

Answer 1

Answer:

Step-by-step explanation:

AB=CD

​  

    ........(A)

∴OR=OQ     ...(3) (equal chords have equal distance from the centre of the circle)

OP  

2

=OR  

2

+RP  

2

      ......(1)

OP  

2

=OQ  

2

+QP  

2

       ....(2)

From (1) , (2) and (3)

OP  

2

=RP  

2

       ....(4)

⇒  

QP+RP

​  

.....(4)

AB=CD

2

AB

​  

=  

2

CD

​  

 

QB=DR

​  

    (perpendicular from centre bisech the chord)

(ii) On adding (4) and (5)  

QB+QP=RP+DR⇒  

BP=DP

​  

     ......(6)

(A)−(6)

(i) ⇒AB−BP=CD−DP

⇒  

AP=CP

​  

     ....(7)

∴  

AP=CP

​  

 &  

BD=DP

​