In the given figure, XY and X?Y? are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, is intersecting XY at A and X?Y? at B. Prove that \[\angle AOB=90{}^\circ \]. |
Answer:
Given, XY & X?Y? are parallel Tangent AB is another tangent which touches the circle at C. To prove: \[\angle AOB=90{}^\circ \] Const.: Join OC. Proof: In \[\Delta \text{ }OPA\]and \[\Delta \text{ }OCA\] \[OP=OC\] (Radii) \[\angle OPA=\angle OCA\] (Radius \[\bot \] tangent) \[OA=OA\] (Common) \[\therefore \] \[\Delta \,OPA\cong \Delta \,OCA\] (CPCT) \[\therefore \] \[\angle 1=\angle 2\] ...(i) Similarly, \[\Delta \,OQB\cong \Delta \,OCB\] \[\therefore \] \[\angle 3=\angle 4\] ...(ii) Also, POQ is a diameter of circle \[\therefore \] \[\angle POQ=180{}^\circ \] (Straight angle) \[\Rightarrow \] \[\angle 1+\angle 2+\angle 3+\angle 4=180{}^\circ \] From eq. (i) and (ii) \[\angle 2+\angle 2+\angle 3+\angle 3=180{}^\circ \] \[2(\angle 2+\angle 3)=180{}^\circ \] \[\angle 2+\angle 3=90{}^\circ \] Hence, \[\angle AOB=90{}^\circ \] Hence Proved.
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