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Suppose that we have a right triangle ABC with the right angle at B such that AC = \(\sqrt{61}\) and AB = 5. A circle is drawn with its center on AB such that the circle is tangent to AC and BC. If P is the point where the circle and side AC meet, then what is CP?

Mathgenius  Oct 6, 2018
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BC  = √ [ AC^2  - AB^2 ] = √ [ 61  - 25 ] = √36  =  6

 

If the center of the circle  is on AB  and is tangent to BC, then  it is tangent to BC at B

And this tangent  drawn from C  = 6  = BC

 

But CP  is also a tangent to the circle drawn from C.....and if P is also tangent to the circle...then P's distance from C must be the same length as  BC  = 6

 

So  CP  = 6

 

Proof :

 

Bisect angle BCA....let this bisector intersect AB  at D

This creates two  triangles,  DBC and DPC

angle CPD  = angle DBC  = 90

angle PCD  = angle BCD  [ by angle bisector CD]

CD  = CD

 

So...by AAS  triangle DBC  = triangle DPC

 

So....BC  =  PC   = 6

 

 

 

cool cool cool

CPhill  Oct 7, 2018

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