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I need a graph of this problem?

In a rhombus, $ABCD$, line segments are drawn within the rhombus, parallel to diagonal $BD$, and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment as a function of its distance from vertex $A$. The graph is:

$\textbf{(A)}\ \text{A straight line passing through the origin.}\\ \textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.}\\ \textbf{(C)}\ \text{Two line segments forming an upright V.}\\ \textbf{(D)}\ \text{Two line segments forming an inverted V.}\\ \textbf{(E)}\ \text{None of these.}$

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closed as off-topic by kingW3, Dando18, José Carlos Santos, Leucippus, Shailesh Sep 17 '17 at 2:07

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  • $\begingroup$ @Doug M help please $\endgroup$ – user373141 Sep 14 '17 at 16:55
  • $\begingroup$ @Arthur , honestly , l completely lost in this problem, please help me understand it $\endgroup$ – user373141 Sep 14 '17 at 17:10
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While segment of line, that is parallel to BD, lets call it B'D', |B'D'| is increasing, while B'D' is inside triangle ABD and decreasing while B'D' is inside triangle BCD |B'D'| is decreasing. So the answer is (D) - inverted V

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  • $\begingroup$ thank you very much, can you elaborate a bit more please, is it possible to provide a sketch for this problem $\endgroup$ – user373141 Sep 14 '17 at 17:32
  • $\begingroup$ Ok, look, lets call B1D1 and B2D2 - two different segments of lines, that are parallel to BD. Let dist(B2D2, A) > dist(B1D1, A). So triangles AB2D2 and AB1D1 are almost similar: AB2 = $alpha * AB1$, AD2 = $alpha * AD1$ and B2D2 = $alpha * B1D1$, and $alpha$> 1 $\endgroup$ – Cheburek Sep 14 '17 at 17:49
  • $\begingroup$ Here B2D2 and B1D1 are inside ABD, almost similar, when they are inside BCD $\endgroup$ – Cheburek Sep 14 '17 at 17:57

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