`X Y`and `Xprime Yprime`are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting `X Y`at A and `XprimeYprime`at B. Prove that `/_A O B\ =\ 90^0`

Explanation

Since, tangents drawn from an external point to a circle are equal.
`:." "AP=AC`
Thus, in `triangleAPO` and `triangleACO,`
`:." "{{:(AP=AC),(AO=AO" ""(common)"),(OP=OC" ""(radius of circle)"):}`
`:.` By SSS cirterion of congruence, we have
`triangleAPO~=triangleACO" "implies" "anglePAO=angleOAC" "`(c.p.c.t.)
`implies" "anglePAC=2angleCAO`
Similarly, we can prove that
`angleCBO=angleOBQ" "implies" "angleCBQ=2angleCBO`
Since, `XY||X'Y'`
`anglePAC+angleQBC=180^(@)`
(sum of interior angles on the same side of transversal is `180^(@)`)
`:." "2angleCAO+2angleCBO=180^(@)" "...(1)`
`implies" "angleCAO+angleCBO=90^(@)`
In `triangleAOB,`
`implies" "angleCAO+angleCBO+angleAOB=180^(@)" "`(angle sum property)
`implies" "angleCAO+angleCBO=180^(@)-angleAOB" "...(2)`
`:.` From Eqs. (1) and (2), we get
`180^(@)-angleAOB=90^(@)" "implies" "angleAOB=90^(@).`
Hence Proved.