### 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

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}
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} 