In the given figure, XYXY and X′Y′ are two parallel tangents to a circle with center O and another tangent AB with the point of contact C intersects XY at A and X′Y′ at B. Prove that ∠AOB=90∘.
Answer:
- We know that tangent at any point is perpendicular to the radius through the point of contact. So, ∠APQ=90∘ and ∠BQP=90∘.(i)
- Consider quadrilateral APQB ∠APQ+∠BQP+∠QBA+∠PAB=360∘[Sum of angles of a quadrilateral is 360∘.]⟹∠QBA+∠PAB=360∘−(∠APQ+∠BQP)⟹∠QBA+∠PAB=360∘−(90∘+90∘)[Using eq (i)]⟹∠QBA+∠PAB=180∘⟹∠QBC+∠PAC=180∘ [As, ∠PAB is same as ∠PAC and ∠QBA is same as ∠QBC] …(ii)
- We know that the tangents from an external point are equally inclined to the line segment joining the center to that point.
So, ∠CAO=12∠PAC and ∠CBO=12∠QBC Therefore, ∠CBO+∠CAO=12(∠QBC+∠PAC)⟹∠CBO+∠CAO=12×180∘=90∘ [Using eq (ii)] ⟹∠ABO+∠BAO=90∘[As,∠CAO is same as ∠BAO and ∠CBO is same as ∠ABO] …(iii) - In △AOB, we have ∠BAO+∠AOB+∠ABO=180∘[Sum of angles of a triangle is 180∘.]⟹∠AOB=180∘−(∠ABO+∠BAO)⟹∠AOB=180∘−90∘=90∘[Using eq (iii)] ⟹∠AOB=90∘