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

enter image description here

  • $\begingroup$ I edited the question $\endgroup$ – AmHsnSharafi Apr 22 '14 at 14:14

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:

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

  • $\begingroup$ Sorry, glanced up at the wrong name while writing my answer. Fixed. $\endgroup$ – Travis Bemrose Apr 22 '14 at 13:02

$\triangle BEA$ 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.

  • $\begingroup$ 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

Denote $AB=a$, $BC=b$, $\dfrac{AB}{BE}=k$.


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

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

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

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

  • $\begingroup$ 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

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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.