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In the figure, $ABCD$ is a convex quadrilateral and E and F are respectively the midpoints of diagonals $AC$ and $BD$.
Prove that:
$$AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2 + (4*EF^2)$$ (Hint: Use the property of the length of the formula of a median in $\triangle BDA$, $\triangle BDC$ and $\triangle AFC$)

$$m_c = \frac{1}{2}\sqrt{2CD^2+2CB^2-DB^2}$$ $$m_a = \frac{1}{2}\sqrt{2AD^2+2AB^2-DB^2}$$ $$m_f = \frac{1}{2}\sqrt{2FC^2+2FA^2-AC^2}$$ $$EF = \frac{1}{2}\sqrt{\sqrt{2CD^2+2CB^2-DB^2}+\sqrt{2AD^2+2AB^2-DB^2}-AC^2}$$ $$4EF^2 = \sqrt{2CD^2+2CB^2-DB^2}+\sqrt{2AD^2+2AB^2-DB^2}-AC^2$$ $$4EF^2 + AC^2 - \sqrt{2CD^2+2CB^2-DB^2} = \sqrt{2AD^2+2AB^2-DB^2}$$ $$(4EF^2 + AC^2 - \sqrt{2CD^2+2CB^2-DB^2})^2 = 2AD^2+2AB^2-DB^2$$ $$(4EF^2+AC^2)^2 - 2(4EF^2 + AC^2)\sqrt{2CD^2+2CB^2-DB^2} +2CD^2+2CB^2-DB^2 = 2AD^2+2AB^2-DB^2 $$ $$(4EF^2+AC^2)^2+2CD^2+2CB^2-2AD^2-2AB^2=2(4EF^2 + AC^2)\sqrt{2CD^2+2CB^2-DB^2} $$ $$((4EF^2+AC^2)^2+2CD^2+2CB^2-2AD^2-2AB^2)^2=4(4EF^2 + AC^2)^2(2CD^2+2CB^2-DB^2)$$

Thank you in advance! img1

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    $\begingroup$ I believe the question's hint is referring to the first set of formulas in Formulas involving the medians' lengths. Have you tried using those formulas? $\endgroup$ Commented Jan 31 at 7:20
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    $\begingroup$ There undoubtedly is more than one method to get the result. Regardless, note that I was able to relatively simply and easily prove the requested result by using those formulas. $\endgroup$ Commented Jan 31 at 7:27
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    $\begingroup$ With all due respect, I will not (although a quick look indicates you forgot to square the various median lengths, which would have removed the square roots). Note that basically all of us, including me, are volunteers, who would like to help, but please make it easier for us. In particular, don't use images for math text (see Using text and MathJax instead of screenshot., and its various linked posts, for several reasons why). Instead, use MathJax (which is similar to $\LaTeX$; see ... $\endgroup$ Commented Jan 31 at 16:23
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    $\begingroup$ (cont.) MathJax basic tutorial and quick reference for details). In your case, note that you surround inline math with single dollar signs, i.e., $, and display equations (i.e., on a separate line with vertical space above & below) with double dollar signs, i.e., $$. For subscripts, fractions, square roots and exponents, e.g., your first line of $m_c=\frac{1}{2}\sqrt{2b^2+2b^2-c^2}$, you can use m_c=\frac{1}{2}\sqrt{2b^2+2b^2-c^2}. Finally, including a solution in your question is basically equivalent to using a solution-verification ... $\endgroup$ Commented Jan 31 at 16:27
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    $\begingroup$ (cont.) tag. As such, you should state explicitly what you're unsure about rather than having somebody try to go through your solution. In your case, it's relatively short and straight-forward but, in general, note we are a question and answer site, for specific questions, rather than more general tutoring. I hope I didn't appear to be harsh here, but I believe you'll get a better response from people here by following guidelines like this, with my just trying to give you what I consider to be useful advice. Also, I feel you would like to, and can, become a valuable member of this community. $\endgroup$ Commented Jan 31 at 16:31

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$$2AF^2 = AD^2 + AB^2 - \frac{DB^2}{2}$$ ,Using the length of median in triangle BDA with AF as Median $$2CF^2 = CD^2 + CB^2 - \frac{DB^2}{2}$$,Using the length of median in triangle BCD with CF as Median $$2EF^2 = AF^2 + FC^2 - \frac{AC^2}{2}$$,Using the length of median in triangle AFC with FE as Median Adding all the equations you get the answer, You should know these median results Refer to this if you want to read more refer to ch3 geometry of this book https://www.ncjindalps.com/pdf/MATHEMATICS/Challenge%20and%20Thrill%20of%20Pre-College%20Mathematics%20-%20V%20Krishnamurty.pdf

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