Triangle with concurrent parallel line segments

[LittleWolf]

I was asked to solve the following problem. A triangle has sides of length 12,8 and 6. There are line segments of equal length that are parallel to each side of the triangle and intersect at one point(concurrent) inside the triangle, what is the length of line segment? I solved the problem by setting up a system of equations. The length of the line segments turned out to be 16/3. Does anyone know how to solve this problem using euclidean geometry?

 

[uart]

You know that unless you place some other restriction (for example that the ends of the line segments must lay on the sides of the triangle for example) then there is no restriction to either the lengths of the line segments or to the position of their intersection point. Are you sure you haven't left something out of the problem description?

 

[tacman]

Yes, this problem can also be solved using Euclidean geometry. Here is one possible approach:

First, draw a diagram of the triangle and the parallel line segments intersecting inside the triangle. Label the point of intersection as P, and the line segments as a, b, and c, with a parallel to the side of length 12, b parallel to the side of length 8, and c parallel to the side of length 6.

Next, draw a line from P perpendicular to side a, and label the intersection point with side a as D. Similarly, draw a line from P perpendicular to side b, and label the intersection point with side b as E. Then, draw a line from P perpendicular to side c, and label the intersection point with side c as F.

Now, we have created three right triangles (PDA, PEB, and PFC) that share a common side length of 12/3, 8/3, and 6/3 respectively. By the Pythagorean theorem, we can set up the following equations:

PD^2 + 12^2 = a^2
PE^2 + 8^2 = b^2
PF^2 + 6^2 = c^2

Since we know that a = b = c (as the line segments are of equal length), we can set these equations equal to each other and solve for a single variable:

PD^2 + 144 = PE^2 + 64
PD^2 + 144 = PF^2 + 36

Subtracting the second equation from the first, we get:

0 = PE^2 - PF^2 + 28

Since PE = PF (as they are both equal to the length of the line segment), we can simplify the equation to:

0 = 28

This is a contradiction, so our assumption that PE = PF must be false. Therefore, PE and PF cannot be equal, and the only way for this to be true is if both PE and PF are equal to 0. This means that the length of the line segment is equal to the length of PD, which we can solve for using the Pythagorean theorem:

PD^2 + 12^2 = (16/3)^2
PD^2 = (16/3)^2 - 144
PD = √(256/9 - 144)
PD = 16/3