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A triangle ABC is inscribed in a circle, and tangent at C to the circle is parallel to the bisector of angle ABC. Find the magnitude of CBD, where D is the point of the intersection of the bisector of angle ABC with AC.
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Answer
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Hint: To solve this question given above, first we will join A and C with the center of the circle. Then we will assume BCX as x. The triangle OBC obtained will be an isosceles triangle. From this triangle, we will derive a relation between BOC and OCB. Now as OC is normal, we will write OCX as 90. Then we will derive a relation between OCB and x with the help of this. With the help of these two relations, we will write BOC in terms of x. Now BOC will be double of CAB because the angle subtended by the chord at center is double the angle subtended at circumference. From here we will get the value of x. Then by using the concept of parallel lines, we will find a relation between x and CBD.

Complete step-by-step answer:
In this question, we are going to assume that the segment ABC is a major segment. We know that if ABC is a major segment then the angle ABC will be less than 90. If our assumption is wrong then the value of angle ABC will be greater than 90 in the end. So, now we are going to do some construction in the circle. First, we will join the points A and B to the center (O) of the circle. We will then consider angle BCX = x. Thus, after construction the circle will look like:
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Now, we know that OB and OC are the radii of the circle. So, they will be equal. So, we can say that triangle OBC is an isosceles triangle. Thus, OCB=OBC. Now we will write the angle BOC in terms of x. We know that, in a triangle, the sum of all the interior angles is 180. Thus, in triangle BOC,
COB+OCB+OBC=180
COB+2OCB=180 (OCB=OBC)
COB=1802OCB - (i)
Since XY is a tangent at C and OC passes through the center, so OC is the normal of the circle. Also angle between tangent and normal is 90. So, we will get,
OCX=90
OCB+BCX=90
OCB=90BCX - (ii)
Now, we will put the value of OCB from (ii) to (i). After doing this, we will get:
COB=1802(90BCX)COB=180180+2BCXCOB=2BCX
COB=2x - (iii) (BCX=x)
We can also see that, from the properties of an inscribed triangle, COB=2CAB - (iv).
From (iii) and (iv), we get:
2CAB=2xCAB=xx=40
Now, we are given that the lines BD and XY are parallel. So, DBC=BCX (according to alternate interior angles)
DBC=xDBC=40CBD=40
Now the ABC=80. It is less than 90. So our assumption was right. ABC is a major segment. Thus,
CBD=40

Note: This question can be solved directly by the alternate segment theorem which says that the angle between the chord and the tangent through one of the end points of the chord is equal to the angle in the alternate segment.
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