In the following figure, PQ = QR, `∠`RQP = 68° , PC and CQ are tangents to the circle with centre O
Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively.
To prove the given question, it is sufficient to prove AB = CD.
For this join OM, ON, OB and OD.
Let the radius of outer and inner circles be R and r respectively.
AB touches the inner circle at M.
AB is a tangent to the inner circle
∴ OM ⊥ AB
⇒ BM =`1/2 `AB
⇒ AB = 2BM
Similarly ON ⊥ CD, and CD = 2DN
Using Pythagoras theorem in ΔOMB and ΔOND
`OB^2 = OM^2 + BM^2 , OD^2 = ON^2 + DM^2`
⇒ `BM= sqrt(R^2 - r^2) ,DN = sqrt(R^2 - r^2`
`AB = 2BM = 2 sqrt(R^2 -r^2) ,CD = 2DN =2 sqrt(R^2 -r^2)`
∴ AB = CD