QR is a chord of the circle with centre O. Two tangents drawn at eh points Q and R intersect each other at the point P. If QM is a diameter, prove that `angle QPR=2 angleRQM[GP-X]`

Explanation

QR is a chord of the circle with centre O. Two tangents drawn at the point Q and R intersect each other at the point P. QM is a diameter.
To prove `: angle QPR =2 angle RQm`
Construction :Let us join O,R.
Proof `:OQ bot PS therefore angle OQP =90^(@)`
Again `OR bot PT, therefore angle ORP =90^(@)`
In quadrilateral `OQPR, angle QOR+angle ORP + angle PQO=360^(@)`
or,` angle QOR +90^(@) +angle RPQ +90^(@)=360^(@)`
or, `angle QOR+ angle RPQ =180^(@) or , angle RPQ =180^(@)-angle QOR `
or, `angle RPQ =angle ROM .........(1) [becauseangle QOR+angle ROM =180^(@)]`
Now, `angle RQM` is the angle in circle and `angleROM` is the central angle produced by the choed RM.
`therefore angle ROM =2 angle RQM` [by theorem -34]
`or, angle QPR=2 angle RQM.` (Proved)