CIRCLE-IN TRIANGLE

Label the vertices of the triangle X, Y, and Z where X is the left vertex of the base, and Z is the right vertex of the base. Y is where the two sides of length 60 meet. Label the center of the circle C. Label the midpoint of the base M. The circle intersects side YZ. Label the point of intersection W.

Since the altitude of the isosceles triangle is divided into a 12:5 ratio by the center of the circle, YC = 12x and CM = 5x (x is a variable).

Draw a radius from C to W. CW is perpendicular to YZ (tangents are perpendicular to radii).

You now have similar triangles MYZ and WYC.

Therefore, YW/YM = YC/YZ.

YW = sqrt(119)x [pythagorean theorem: (5x)^2 + YW^2 = (12x)^2]
YM = 17x
YC = 12x
YZ = 60 (given)

Rearrange terms and simplify to get:
17x = 5*sqrt(119)

17x is the altitude of the MYZ, so YM = 5*sqrt(119)

Use the pythagorean theorem [MZ^2 + YM^2 = 60^2] to get:
MZ = 25

XZ = 2*MZ = 50

Hello!!
Thanks Markr for your thoroughly given answer. Just want you to know
that I` been asking another math forum index for help and hope that you
will not be offended(please do not be).
Thank you so very much I` really appreciate.

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