triangles, midpoints and parallelism

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

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Re: triangles, midpoints and parallelism

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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 = $\displaystyle \alpha$ AB, BE = $\displaystyle \beta$ AB where $\displaystyle \alpha , \beta$ are some constants. Using the condition $\displaystyle AD \le AE$ we have $\displaystyle \alpha + \beta \le 1$.

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 $\displaystyle 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), $

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

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

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

Kalyan

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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.

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Re: triangles, midpoints and parallelism

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

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