In the given figure, ^@ XY ^@ 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 ^@ \angle AOB = 90^\circ^@.

O Q P C B A X X' Y' Y


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Answer:


Step by Step Explanation:
  1. We know that tangent at any point is perpendicular to the radius through the point of contact. @^ \begin{aligned} \text{ So, } \angle APQ = 90^\circ \text{ and } \angle BQP = 90^\circ. && \text{(i)} \end{aligned} @^
  2. Consider quadrilateral ^@ APQB ^@ ^@ \begin{aligned} &\angle APQ + \angle BQP + \angle QBA + \angle PAB = 360^\circ && \text{[Sum of angles of a } \\ & &&\text {quadrilateral is } 360^\circ.] \\ \implies &\angle QBA + \angle PAB = 360^\circ - (\angle APQ + \angle BQP) \\ \implies &\angle QBA + \angle PAB = 360^\circ - (90^\circ + 90^\circ) && \text{[Using } eq \text { (i)]} \\ \implies &\angle QBA + \angle PAB = 180^\circ \\ \implies &\angle QBC + \angle PAC = 180^\circ && \text{ [As, } \angle PAB \text { is same as } \angle PAC \text { and } \space \\ & && \angle QBA \text{ is same as } \angle QBC] \space \ldots \text{(ii)} \end{aligned} ^@
  3. We know that the tangents from an external point are equally inclined to the line segment joining the center to that point.

    So, @^ \angle CAO = \dfrac { 1 } { 2 } \angle PAC \text{ and } \angle CBO = \dfrac { 1 } { 2 } \angle QBC @^ Therefore, ^@ \begin{aligned} &\angle CBO + \angle CAO = \dfrac { 1 } { 2 } ( \angle QBC + \angle PAC) \\ \implies &\angle CBO + \angle CAO = \dfrac { 1 } { 2 } \times 180^\circ = 90^\circ && \text{ [Using } eq \text{ (ii)] } \\ \implies &\angle ABO + \angle BAO = 90^\circ && { [As, } \angle CAO \text{ is same as } \angle BAO \text{ and } \\ & && \angle CBO \text{ is same as } \angle ABO ] \space \ldots \text{(iii)} \end{aligned} ^@
  4. In ^@ \triangle AOB ^@, we have ^@ \begin{aligned} &\angle BAO + \angle AOB + \angle ABO = 180^\circ && \text{[Sum of angles of a triangle is } 180^\circ.] \\ \implies &\angle AOB = 180^\circ- ( \angle ABO + \angle BAO ) \\ \implies &\angle AOB = 180^\circ - 90^\circ = 90^\circ && \text{[Using } eq \text{ (iii)] } \\ \implies &\angle AOB = 90^\circ \end{aligned} ^@

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