Mathematics
Two circles with centers O and O' are drawn to intersect each other at points A and B. Center O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with center O' at A. Prove that OA bisects angle BAC.
Answer
Join O'A and O'B.
As, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :
CD is the tangent and AO is the chord.
∴ ∠OAC = ∠OBA ……… (1)
In ∆OAB,
OA = OB [Radius of the circle with center O.]
∠OAB = ∠OBA ………. (2) [Angles opposite to equal sides]
From (1) and (2), we have
∠OAC = ∠OAB
Hence, proved that OA is the bisector of ∠BAC.
Related Questions
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠BCG = 108° and O is the center of the circle, find :
(i) angle BCT
(ii) angle DOC
If PQ is a tangent to the circle at R; calculate :
(i) ∠PRS,
(ii) ∠ROT.
Given, O is the center of the circle and angle TRQ = 30°.
In the figure (i) given below, PQ is a tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate
(i) ∠QAB
(ii) ∠PAD
(iii) ∠CDB.
