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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 13 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 13 hours ago
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    $\begingroup$ also you should show what you tried. $\endgroup$ Commented 13 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 13 hours ago
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    $\begingroup$ @Intelligentipauca. .very nice trigonometric solution. $\endgroup$ Commented 9 hours ago

3 Answers 3

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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 13 hours ago
  • $\begingroup$ awesome as always.. $\rlap \smile {\dot{}\dot{}}$ $\endgroup$ Commented 13 hours ago
  • $\begingroup$ Yes... Excellent.[+1]. Thanks. $\endgroup$ Commented 12 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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The line segment $FG$ is inconveniently placed to compare it to $BC$. I would much rather compare $BC$ to another chord of the circle. In the diagram, it sure looks like there is a chord $EH$ parallel and congruent to $FG$, so let's try to prove that.

Extend line $DG$ so that it intersects the circle at $H$. Then $\angle EFG$ and $\angle FGH$ are right angles because $EF$ and $DG$ are perpendicular to $AB$, and $\angle GHE$ is a right angle because it subtends a diameter, so $EFGH$ is a rectangle. In particular, $EH = FG$.

The diagram in the proof with H added as in the text

The chords $AB$ and $EH$ are subtended by $\angle C$ and $\angle D$, respectively, so to prove that the chords are equal, it's enough to prove that $\angle C = \angle D$. But once we set ourselves such a goal, there are lots of ways to prove this by a bit of angle chasing. For example, if $DE$ intersects $AC$ at $X$ and $BC$ at $Y$, then $\triangle CXY$ and $\triangle DGY$ both have a right angle ($\angle CXY$ and $\angle DGY$, respectively) and share $\angle CYX = \angle DYG$, so their third angle is also equal: $\angle XCY = \angle GDY$.

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