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Let triangle ABC be an isosceles triangle with AB=AC. Let D and E be the midpoints of AB and BC respectively. Given that there exists a point F on ray DE outside of triangle ABC such that triangle BFA is similar to triangle ABC, compute AB/BC.

I did this and got 2 but the answer given is √2. Solution?

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  • $\begingroup$ What did you do? $\endgroup$ – Pakquebchsoflwty Oct 26 '14 at 23:13
  • $\begingroup$ I made AB and AC x. Using the similar triangles given I solved for BF as X too, however I'm not sure where to go from here. I don't remember exactly what I did but the answer i arrived at was incorrect. $\endgroup$ – Lulu Uy Oct 26 '14 at 23:18
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enter image description here

First of all, triangle BFA has to be congruent to triangle ABC. Therefore, all the pink lines in Fig. 1 are equal. This further leads to the dark blue lines and light blue lines are equal in length in Fig. 2.

Also, in Fig.1 and fig. 2, hope you understand why all same color marked angles are equal. Note also that (pink angle) - (green angle) = (yellow angle).

Fig. 3 is a side track to the midpoint theorem. From which, we have the shaded quadrilateral is a //gm with DE = EF = 0.5BC.

The two triangles shown in Fig. 4 are similar. The corresponding ratios give the required answer.

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  • $\begingroup$ Seriously why don't you download GeoGebra? :) $\endgroup$ – Sawarnik Oct 27 '14 at 18:11
  • $\begingroup$ @Sawarnik Thankyou for the advice. I do have GeoGebra. Sometimes (probably because of habit matter or because I am not that good in using the s/w) I found PC paintbrush is easier for drawing simple figures. I use GeoGebra only when I need the corresponding equations or angles with accurate measures etc. Even the figure was drawn in GeoGebra, I still need to send it to PC paintbrush for the finishing touch. Sorry again for the not-so-pretty diagrams but I hope they can do the explanation job. $\endgroup$ – Mick Oct 28 '14 at 3:47
  • $\begingroup$ Well, as you feel comfortable. Like habit has made me more comfortable with GeoGebra (accurate and quick). But I don't understand what do you mean by the finishing touch, is it the colour, line style, etc? That can be easily done by right clicking to object properties and you have all the options. :) $\endgroup$ – Sawarnik Oct 28 '14 at 9:10
  • $\begingroup$ @Sawarnik By finishing touch, I mean erasing some of the unnecessary parts (like part of a ray instead of segment), labelling/renaming a point or a line etc. Anyway, will use that tool more often. $\endgroup$ – Mick Oct 28 '14 at 15:46
  • $\begingroup$ That can be easily done too, just right click to Show Object, and poof .. and you have them their at Algebra toolbar. The little button at the beginning of an object in the Algebra toolbar is also the same thing. :) $\endgroup$ – Sawarnik Oct 28 '14 at 17:49

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