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

#1**+1 **

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

CPhill
Oct 7, 2018