Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: AB is tangent to a circle with centre O.
To prove: OP is perpendicular to AB.
Constructions: Take a point Q on AB and join OQ.
Proof: Since Q is a point on the tangent AB, other than the point of contact P, so Q will be outside the circle.
Let OQ intersect the circle at R.
Now OQ = OR + RQ
Thus OP is shorter than any other segment among all and the shortest length is the perpendicular from O on AB.
Let ABCD be a quadrilateral, circumscribing a circle.
Since the tangents drawn to the circle from an external point are equal,
AP = AS ..........(1)
BP = BQ ..........(2)
RC = QC ..........(3)
DR = DS ..........(4)
Adding, (1), (2), (3) and (4), we get
AP + PB + RC + DR = AS + BQ + QC + DS
(AP + PB ) + (RC + DR ) = (AS + DS ) + (BQ + QC)
AB + CD = AD + BC.