# triangles, midpoints and parallelism

• Jan 9th 2013, 04:38 AM
mathmagix
triangles, midpoints and parallelism
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 $MP\parallel AB$.
• Jan 11th 2013, 09:36 AM
kalyanram
Re: triangles, midpoints and parallelism
Attachment 26533
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
• Jan 11th 2013, 07:11 PM
johng
Re: triangles, midpoints and parallelism
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.
Attachment 26540
• Jan 13th 2013, 09:54 PM
LoidaWard
Re: triangles, midpoints and parallelism
As implied above, MP is parallel to AB iff length AD = length EB.

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