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.
Circles
8 Likes
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.
Answered By
5 Likes
Related Questions
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.
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, 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
In the figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that :
∠CAD = [∠PBA - ∠PAB]
![In the figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that ∠CAD = 1/2 [∠PBA - ∠PAB]. Tangents and Intersecting Chords, Concise Mathematics Solutions ICSE Class 10.](https://cdn1.knowledgeboat.com/img/cm10/q12-c18-ex-18-b-tangents-concise-maths-solutions-icse-class-10-1200x800.png)