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I found this problem in a French paper translated from Arabic in 1927 by an author named Al Bayrouni . I wonder if it can be found in one of Archimedes' works . Here is the statement :

ABC is a triangle, the perpendicular bisector of [AC] intersects the circumcircle of ABC at D and E . Let F and G be the orthogonal projections of E and D onto (BC) , then : |FG|=|AB|

In any case , this theorem can be proven using complexe numbers for example ( which is what I did ) , but I wonder how to do it more simply by the synthetic geometry !? enter image description here

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    $\begingroup$ Al Baruni was a persian polymath. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ also why don't you use geoggebra classic to draw your digram they would look neater and better (just a suggestion) $\endgroup$ Commented 10 hours ago
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    $\begingroup$ also you should show what you tried. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ "I wonder if it can be found in one of Archimedes' works" The folks at History of Science and Math SE may be able to address that. $\endgroup$ Commented 10 hours ago
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    $\begingroup$ @Intelligentipauca. .very nice trigonometric solution. $\endgroup$ Commented 5 hours ago

2 Answers 2

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enter image description here

Hint: In the picture $GH||BI$. You have to show:

1-$HI||BG$

2- Circle $c_1$ is tangent to $BC$ at $K$ and to $AB$ at $A$.

3-$KI||FH$

In this way:

$\triangle GFH=\triangle ABI \Rightarrow AB=FG$

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  • $\begingroup$ great answser ....... +1 $\endgroup$ Commented 9 hours ago
  • $\begingroup$ awesome as always.. $\rlap \smile {\dot{}\dot{}}$ $\endgroup$ Commented 9 hours ago
  • $\begingroup$ Yes... Excellent.[+1]. Thanks. $\endgroup$ Commented 9 hours ago
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A simple proof using trig:

$AB = 2R \sin \angle C$.

Let $X$ be the intersection of $BC$ and $DE$ and $O$ be the center of the circumcircle.

Then $GX=(R+OX)\cos(90^\circ-\angle C)$, $XF = (R-OX)\cos(90^\circ-\angle C)$

Adding it together, $GF = 2R \sin \angle C$

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