Expert: Sherry Wallin - 7/6/2009

Question
Hi,
I've got a maths problem:
Let ABC be a triangle. Two straight lines which are parallel to the side AC divide the triangle into three figures of equal area. In what parts is the side AB divided by the lines, if AB = 10 cm?

Cheers.

Answer
Hi James~
    Your question is not clear to me. What do you mean by 'what parts is the side AB divided by the lines,'? Do you mean what proportion? Assuming you mean proportions, take the triangle ABC and draw two straight lines parallel to the side AC so that AB is divided into pieces x1, x2, x3. Notice x1 + x2 + x3 = AB = 10cm. The uppermost parallel line call b1 and the 2nd parallel line call b2 and call the side opposite angle B, b. Using proportions for similar triangles we get:
x1/b1 = * {[x1 + x2]/b2} = 10/b. Also b1/b2 = x1/x2 -> [b1/b2]x2 = x1 and x2 = [b2/b1]x1. Now substitute into * for x1 and x2 which will give you x1 = [10b2]/[b(1+(b2/b1))] and x2 = [10b2]/[b(1+(b1/b2))]. Remember that AB = 10cm and that x1 + x2 + x3 = 10 cm so x3 = 10 - x1 - x2 or
x3 = 10 - [10b2]/[b(1+(b2/b1))] - [10b2]/[b(1+(b1/b2))].  You now have the proportions for any two lines parallel to AC.

This is kind of messy but it is accurate and complete. Please let me know if you need anything explained further.

Math Prof