Suppose that we are given triangle ABC is isosceles. Known length its sides and base. Need to find the median, descended on the base of this triangle. In an isosceles triangle the median is also the median, bisector and height. Due to this property, find the median to the base of the triangle is very simple. Use Pythagorean theorem for the right triangle ABD: AB2 = BD2 + AD2, where BD is the desired median, AB - side (for convenience, let it equal a), and AD is half of the base (for convenience, take the basis equal to b). Then BD2 = a2 - b2/4. Find the root of this expression and get the length of the median.
Slightly more complicated is the situation with the median drawn to the lateral side. To start, draw both of these medians in the figure. These medians are equal. Label the side with the letter a, and the base is b. Mark equal angles at the base α. Each median divides the side into two equal parts a/2. Mark the desired length of the median x.
Cosine theorem to Express any side of a triangle using the other two and the cosine of the angle between them. We can write the theorem of cosines for the triangle AEC: AE2 = AC2 + CE2 - 2AC·CE·cos∠ACE. Or what is the same, (3x)2 = (a/2)2 + b2 - 2·ab/2·cosα = a2/4 + b2 - ab·cosα. For this task the well-known, but the angles at the base there, so the calculation continues.
Now apply the theorem of cosines to triangle ABC to find the angle at the base: AB2 = AC2 + BC2 - 2AC·BC·cos∠ACB. In other words, a2 = a2 + b2 - 2ab·cosα. Then cosα = b/(2a). Substitute this expression into the previous one: x2 = a2/4 + b2 - ab·cosα = a2/4 + b2 - ab·b/(2a) = a2/4 + b2 - b2/2 = (a2+2b2)/4. Calculating the root of the right side of the expression, you'll find the median drawn to the lateral side.