Let's begin by understanding this challenging geometry problem. We have a triangle ABC with a special point D. The angle at vertex A is divided by segment AD into two equal angles of 60 degrees each. We're given that AB is twice the length of AC, and AD has a length of 100. Our task is to find the length of side BC.Now let's set up our variables. We'll denote the length of AC as x. Since AB is twice AC, we have AB equals 2x. We know AD is 100, and both angle BAD and angle CAD are 60 degrees. With these variables established, we can apply the Law of Cosines to triangles ABD and ACD to find relationships between the sides.Now we apply the Law of Cosines to both triangles. In triangle ABD, we have BD squared equals AB squared plus AD squared minus 2 times AB times AD times cosine of 60 degrees. Substituting our values, this becomes 4x squared plus 10000 minus 200x. Similarly, in triangle ACD, CD squared equals x squared plus 10000 minus 100x. These two equations will help us find the value of x.To find BC, we need another equation. Let's apply the Law of Cosines to triangle BCD. First, notice that angle BAC equals 120 degrees, since it's the sum of the two 60-degree angles. To use the Law of Cosines on triangle BCD, we need angle BDC. We can find this using angle relationships in the configuration. Once we have angle BDC, we can write BC squared in terms of BD, CD, and the cosine of angle BDC.To find angle BDC, we use the Law of Cosines again. In triangle ABD, we can express cosine of angle ADB in terms of the sides. Similarly, in triangle ACD, we find cosine of angle ADC. Since angles ADB and ADC are on opposite sides of line AD, angle BDC equals 360 degrees minus angle ADB minus angle ADC. This gives us the angle we need for our final calculation.