#### Question

In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.

#### Solution

i) PQ = RQ

∴ `∠`PRQ = `∠`QPR (opposite angles of equal sides of a triangle)

⇒ `∠`PRQ + `∠`QPR + 68° = 180°

⇒ `∠`2 PRQ = 180° - 68°

⇒ `∠`PRQ =`(112°)/2 = 56°`

Now, `∠`QOP = 2`∠` PRQ (angle at the centre is double)

⇒ QOP = 2 × 56° = 112°

ii) `∠`PQC = `∠`PRQ (angles in alternate segments are equal)

`∠`QPC = `∠`PRQ (angles in alternate segments)

∴ `∠`PQC = `∠`QPC = 56° (∵ `∠`PRQ =56° from(i))

`∠`PQC + `∠`QPC + `∠`QCP = 180°

⇒56° + 56° + `∠`QCP = 180°

⇒ `∠`QCP = 68°

Is there an error in this question or solution?

Solution In Two Concentric Circles, Prove that All Chords of the Outer Circle, Which Touch the Inner Circle, Are of Equal Length. Concept: Arc and Chord Properties - If Two Chords Intersect Internally Or Externally Then the Product of the Lengths of the Segments Are Equal.