+188 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?
+128475 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
