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The problem is as follows:

The figure from below shows a lactose crystal which is about to be studied for its refraction index, for this purpose an orange light beam is focused by means of a lens over a LED. The crystal was cut as a thin plate. The plate is a rectangle $ABCD$. The measured $AB=15\overset{\circ}{A}$ and $FH= 6\,A^{\circ}$. Using this information find the distance $DE$ in Angstrom.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&3\overset{\circ}{A}\\ 2.&5\overset{\circ}{A}\\ 3.&4\overset{\circ}{A}\\ 4.&6\overset{\circ}{A}\\ \end{array}$

Although it is not specified in the problem. I'm assuming that the path which will take the light will be on all the blue lines of the figure, which it would make this into a find the distance $DE$ problem.

However I'm not very sure on how to find such distance. What I thought to do was to use similarity between the triangles:

$$\triangle ABC ~ \triangle FHC$$

But I am still stuck with that part. I have no idea how to continue or what to do then. Can someone help me here?.

I think there may be needed some sort of construction or something, but I don't know. Does it exist a way to solve this relying only in Euclidean postulates?.

Please include a drawing in the answer this part is important because I don't know very well how to spot the right relations here to find the requested distance.

Chris Steinbeck Bell
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    Chris, you have asked $243$ questions on this website till date and nearly all of them have received answers. Yet you have only accepted $14$ solutions in totality. I can't accept that most of the answers you received were not satisfactory for you. You do realize that gaining reputation through answers is the incentive for users to share knowledge on this website. I strongly recommend that you go through all your questions and accept satisfactory solutions to reward people who take their time out to solve your problems. You just need to press the tick-mark button next to their answers. – Shubham Johri Oct 30 '20 at 06:18
  • Seconded. Let's reward our answerers! :-) – Brian Tung Oct 30 '20 at 06:33

1 Answers1

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$\triangle CAB\sim\triangle CHF$ so $CH/CA=FH/AB=2/5$.

$\triangle AHF\sim\triangle ACE$ so $FH/CE=(AC-CH)/AC=1-2/5=3/5$.

This gives $CE=5FH/3=10$ and $DE=CD-CE=15-10=5.$

Shubham Johri
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