question_answer

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.