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We have $AH=HB$ and $BG=GC$ in the image below. Why is $AD=2\times FG$?

enter image description here

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  • $\begingroup$ I edited the question $\endgroup$ – AmHsnSharafi Apr 22 '14 at 14:14
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It's not. As MvG points out, you've got similar triangles, but the homothety he correctly identifies, does not have scale factor $2$. In order for the scale factor to be $2$, one would need $G \mapsto C$ and $H \mapsto A$.

As ratios:

$$\frac{AD}{KF} = \frac{AB}{KB} \ne \frac{AB}{HB} = \frac{2}{1}.$$

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  • $\begingroup$ Sorry, glanced up at the wrong name while writing my answer. Fixed. $\endgroup$ – Travis Bemrose Apr 22 '14 at 13:02
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$\triangle BEA$ is similar to $\triangle BGK$ due to the right angle and the common angle at $B$. For the same reason, $\triangle BCI$ is similar to $\triangle BJH$. So the map characterized by

$$B\mapsto B\quad J\mapsto C\quad K\mapsto A\quad G\mapsto E\quad H\mapsto I\quad F\mapsto D$$

might be a single homothety. To make sure that it actually is, you'd have to verify that the scale factors for the two triangle similarities I mentioned actually agrees. Or equivalently that the points $B,D,F$ are on a single line.

When I wrote this answer, I thought that the scale factor was obviously $2$, but that was because I mixed up some points. I'll leave this as an incomplete answer for now since others refer to it.

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  • $\begingroup$ Why the scale factor is 2? $\endgroup$ – AmHsnSharafi Apr 22 '14 at 13:02
  • $\begingroup$ @AmHsnSharafi: It is not, sorry. I had been to quick to assume $AH=HB\;\Rightarrow AB=2HB$, but to conclude that scale factor, you'd need $IB=2HB$. My mistake for reading this too quickly. $\endgroup$ – MvG Apr 22 '14 at 13:04
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Denote $AB=a$, $BC=b$, $\dfrac{AB}{BE}=k$.

Then

$BI = \dfrac{b}{k}$ $\implies$ $AI=AB-BI=a-\dfrac{b}{k}$;

$BJ = \dfrac{a}{2}\cdot k$ $\implies$ $GJ=BJ-GC=\dfrac{ak}{2}-\dfrac{b}{2}$.

$\triangle ADI \sim \triangle JFG$, so

$$ \dfrac{AD}{FG} = \dfrac{k\cdot DI}{FG} = k \cdot \dfrac{AI}{JG}=2k\times \dfrac{a-b/k}{ak-b} = 2. $$

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  • $\begingroup$ you are right so I edited it $\endgroup$ – AmHsnSharafi Apr 22 '14 at 14:19
  • $\begingroup$ I edited too :) $\endgroup$ – Oleg567 Apr 22 '14 at 14:20
  • $\begingroup$ You should change AE by BE in first line. Thancks $\endgroup$ – AmHsnSharafi Apr 22 '14 at 14:48
  • $\begingroup$ @AmHsnSharafi, yes, of course; thank you for accurate reading. $\endgroup$ – Oleg567 Apr 22 '14 at 14:56
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EDIT: Answer updated to reflect the correction of the problem statement.

Draw a line through $G$ perpendicular to $AB$, and a line through $H$ perpendicular to $BC$. Let $L$ be the intersection of these two lines. You can then show that the figure $BHLG$ is proportional to $BADC$, and that the scale factor is $1:2$. Therefore $AD = 2\times HL$.

But $HLGF$ is a parallelogram, and so $HL = FG$, from which your answer follows.

[Remnants of old answer below, because some of the comments make sense only in regard to this answer:]

The [previous] statement [that $AD = 2\times FG$] is not true in general. For counterexamples, try letting $\triangle ABC$ be an equilateral triangle, or let $\angle ACB$ be a right angle.

There may be information missing from the problem statement, perhaps something to do with the circumscribed circle. The fact that there is a circumscribed circle tells us nothing--every triangle has one--so perhaps there is something else we are supposed to know about that circle?

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  • $\begingroup$ Perhaps you are right. I faced to this problem when I was reading a proof of the nine point circle theorem. $\endgroup$ – AmHsnSharafi Apr 22 '14 at 13:22
  • $\begingroup$ But I don't find any counter when I get $\triangle ABC$ to be an equilateral, because concurrent of perpendicular bisectors coincides to concurrent of medians and as you know this concurrent divides medians 2 to 1.Even there is no counter when $\angle ACB$ is right angle. $\endgroup$ – AmHsnSharafi Apr 22 '14 at 13:56
  • $\begingroup$ I see where we got confused. These were counterexamples for the previous statement that $AD=2\times KF$ (which was the question I read), prior to the correction of the question. Replace $KF$ by $FG$ and it is an entirely different problem. $\endgroup$ – David K Apr 22 '14 at 18:07

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