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I just figured out this information for this problem, and I wanted help completing it. The problem is:

 

Let AB be a diameter of a circle, and let C be a point on the circle such that AC = 8 and BC = 6. The angle bisector of angle ACB intersects the circle at point M.Find CM.

 

 

 

Label the point of intersection of BA and CM as P.

By the angle bisector theorem which states:

 

"If a ray bisects an angle of a triangle, then it divides the opposite side of the triangle into segments that are proportional to the other two sides."

 

Then,  AP and BP are proportional to 8:6

 

We can find that AP is 407/, while BP is 30/7. 

 

What now? Thanks!

 

Noori

 Aug 11, 2020
 #1
avatar+1332 
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Hi, Noori! I'm gonna use the angle bisector theorem to solve this simple question. 

 

1/   AB = MR = 10                (AB and MR are diameters of a circle O)  AB ⊥ MR

 

2/   AP : PB = 8 : 6            AP = 8( 10 / 14 ) = 5.714285714

 

3/   PO = AP - AO = 0.714428714

 

4/   Triangle PMO is the right-angled triangle.

 

5/   Angle PMO = arctan(PO / OM) = 8.130102354º

 

6/   And finally,   CM = cos(PMO) * MR = 9.899494937   or  7√2   smiley

 

 Aug 11, 2020
 #2
avatar+189 
0

Hi Dragan! Thank you so much for the response. The steps were very detailed and helped me understand how to solve it. Thank you for the diagram as well!

Noori  Aug 12, 2020

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