triangles, midpoints and parallelism

**mathmagix** · January 9th 2013, 04:38 AM

Let $ABC$ be an triangle. On $AB$ side of the triangle let $D, E$ be two points so that $AD\leq AE$ . Let $DF \parallel BC,F\in AC$ and $EG\parallel AC,G\in BC$ . Let $L , M , N , P , R$ be the midpoints of the segments: $[CF] , [FD] , [DE] , [EG] , [GC]$ . Proof that $[CN] , [PL] , [MR]$ can be the three sides of a triangle. Find a condition so that $MP\parallel AB$ .

**kalyanram** · January 11th 2013, 09:36 AM

triangles, midpoints and parallelism-tri.png

With out any loss of generality we consider an oblique co-ordinate system with the origin as A with B and C along x and y axis such that BC=a, AC=b, AB=c the lengths of the given triangle. Now say AD = $\alpha$ AB, BE = $\beta$ AB where $\alpha , \beta$ are some constants. Using the condition $AD \le AE$ we have $\alpha + \beta \le 1$ .

Using Basic Proportionality Theorem in this triangle and the formula for internal division of a line segment according to a given ratio. We can determine the co-ordinates of the points as $F(0,\alpha b), L(0,\frac{\alpha+1}{2}b), R(\frac{1- \beta}{2}c, \frac{1+\beta}{2}b), G((1-\beta )c, \beta b), D(\alpha c, 0),$

$E((1-\beta)c,0), M(\frac{\alpha}{2} c, \frac{\alpha}{2} b), N(\frac{1- \beta + \alpha}{2}c,0), P((1-\beta)c, \frac{\beta}{2}b)$ .

Use the Cosine Rule to get the inclination between the co-ordinate axes say $\theta$ as $a^2 = c^2 + b^2 - 2bc \cos \theta$ .Now use the distance formula in the oblique system to determine the lengths $MR, CN, PL$ and show that the sum of lengths of any two sides is greater than the third side so they can form a triangle. $\alpha + \beta \le 1$ will prove to be helpful.

Use the slope equation to get the condition between $\alpha , \beta$ for $MP \parallel AB$ .

Kalyan

**johng** · January 11th 2013, 07:11 PM

This partial solution is similar to the above, but the algebra for the first question was beyond me. As implied above, MP is parallel to AB iff length AD = length EB. I put the given figure in a dynamic geometry program which was very helpful. Here's the image.

triangles, midpoints and parallelism-mhftriangle1.png

**LoidaWard** · January 13th 2013, 09:54 PM

As implied above, MP is parallel to AB iff length AD = length EB.

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