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Can anyone prove the angle $\theta$ (smaller angle) between diagonals of a parallelogram is given by the equation

$$\cos(\theta)=\frac{a-c}{a+c}$$

where $2a$ is the base length of parallelogram; $2c$ is the horizontal distance to the edge when we draw a vertical line as seen in the picture below.

enter image description here

I have been trying to get this but haven't had a success. Thanks for your help in advance.

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  • $\begingroup$ what exactly have you tried? $\endgroup$ – danimal Dec 14 '15 at 11:40
  • $\begingroup$ I tried to get everything in terms of angles of side edges but I can't really get to the (a-c)/(a+c), which seems an easy expression, I tried using law of cos and sin at some point too, but no success. Any suggestions? $\endgroup$ – Baha Dec 14 '15 at 11:44
  • $\begingroup$ can you find or construct a triangle with hypotenuse $a+c$ and adjacent $a-c$? $\endgroup$ – danimal Dec 14 '15 at 11:49
  • $\begingroup$ I tried what you said to get a right triangle of (a+c) hypotenuse n adjacent (a-c) on the sketch, but I couldn't get any. $\endgroup$ – Baha Dec 14 '15 at 12:03
  • $\begingroup$ Hi again everyone, I tried to get the answer to my question finally using vector notations and dot product, which is equivalent to law of cosines at the end and I realized that I am probably misunderstanding the question. I better post the problem from zero without any assumption so that it won't get messed up. Is it better to create a new post or do you know if I can modify this post? $\endgroup$ – Baha Dec 16 '15 at 12:26

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