Question

Two circles with centres O and P intersect each other at A and B as shown in following fig. Two straight lines MAN and RBQ are drawn parallel to OP.
Prove that (i) MN = 20 P (ii) MN= RQ.

Answer 1

Given: Two cirdes with centres 0 and P, and MN II OP || RQ 

To prove: (i) MN = 20P  (ii) MN= RQ. 

Construction: OX ⊥ MN,  PY ⊥ MN, OW ⊥  RZ, PZ ⊥ RQ

Proof: Since each angle of the quadrilateral XYZW is a right angle, sc XYZW is a rectangle. 

Also, XYPO is a rectangle.        ...(1) 

Now, perpendicular drawn from the centre to the chord bisects the chord. 

Therefore, MA = 2 XA and AN = 2 AY   ...(2)

Now, MN = MA + AN = 2 XA + 2 AY   [from (2)]

⇒ MN = 2(XA + AY) = 2 XY 

⇒ MN = 2 OP    [As XYPO is a rectangle, XY = OP]   ... (3) 

This proves part (i).

By similar arguments, we have RQ = 2 OP   ...(4)

Using (3) and ( 4), we get 

MN= RQ. 

This proves part (ii).