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
19 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
13 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)