How can I find the angle between two intersecting segments coming from a trapezoid? I'm confused about this problem, since I don't know which Identity or theorem can be used to solve it. I tried to pull all the tricks under my sleeve but yet I can't find an answer, can somebody help me with this? The following figure represents a sugar whose atoms are labeled as $A$, $B$, $C$, $D$, $E$, $F$. $AE = 5 \sqrt{2} \, \overset{\circ}{A}$ , $EF = 5 \sqrt{3} \, FD = 5\, \overset{\circ}{A}$. If it is known that $BC=AB+CD$ and $\angle EBC = \angle FCB = 90^{\circ}$. Calculate the angle formed by the extension of the lines $AE$ and $FD$.See the figure from below which corresponds to the problem: https://i.imgur.com/0K69oRr.png The existing alternatives are: $\begin{array}{ll} 1.& 60^{\circ}\\ 2.& 53^{\circ}\\ 3.& 75^{\circ}\\ 4.& 90^{\circ}\\ 5.& 105^{\circ}\\ \end{array}$ So far what I could come up with was that what it was intended to be found is the angle which is colored by a blueberry shade. https://i.imgur.com/UIgIarL.png But I don't know how to relate it with the known information, hence what can be done?. If possible I would appreciate that the proposed answer could come up with some drawing or sketch as to better visualize what's trying to be meant. I have not posted it 
Place point G on segment BC such that AB = BG. Since AB + CD = BC, then GC = CD. Connect both E and F to G forming two isosceles triangles, AEG and GFD. Note triangle EGF has $EG^2 + GF^2 = EF^2$ See what you can do from here ... 
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Trying to make up a similar problem where triangle EFG is a 345 right triangle; FG=3, EG=4, EF=5. Any tips? Thanks. 
Duhhhh...works for any right triangle with sides similarly given... With 345, fits inside isosceles right triangle with equal sides = 7. Thus solution always 90 degrees. 
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Dan 
Just there but I still need a bit of help Quote:
See figure for more details: https://i.imgur.com/ID4cSOAl.png $\left(5 \sqrt{3}\right)^{2}= \left(5 \sqrt{2}\right)^{2} + \left(5 \right)^{2}  2 \left(5^{2} \sqrt{2}\right)\left(\sqrt{3}\right)\cos\angle EGF$ From this is reduced to: $ 2 \left(5^{2} \sqrt{2}\right)\left(\sqrt{3}\right) \cos\angle EGF = 0$ Therefore and assuming the angle $0 \leq \angle EGF \leq \frac{\pi}{2}$ then: $\angle EGF = 90^{\circ}$ Now the thing which I am still stuck at is how to prove or justify that the $\angle EHF = 90^{\circ}$ for now I am guessing it is going to be that value but I don't know how to establish that. I've tried to use all sorts of lines and parallel lines to do so but still I can't find a way. Can you further help me?. :help: 
Pythagoras? Since angle EGF = 90, then angle EGB and angle FGC are complementary $\implies$ angle A and angle D are complementary $\implies$ angle H = 90. 
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It was after looking to the figure several times and prior reading at your answer that the only thing that I could come up with to justify the $90^{\circ}$ comes by drawing a parallel line to $AD$ so that interior angles summed up with the one which is being asked equate to a right angle. To reflect this I made a new drawing. https://i.imgur.com/VdudOVAl.png Therefore from the figure: $\omega+\phi+\psi=180^{\circ}$ $\psi=180\left(\omega+\phi\right)=18090=90^{\circ}$ Again I'm not sure if this was what you meant but so far I cannot find other explanation. Did I do well?. :) 
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Recommend you make yourself familiar with using the converse of the Pythagorean Theorem. With it one may determine if the longest side of a triangle is opposite a right angle, an obtuse angle, or an acute angle. If not a right angle, then the law of cosines is appropriate. I used the converse to determine $\angle{EGF} = 90$ ... not trial & error. Quote:
angle A and angle D are complementary $\implies \angle{A} + \angle{D}= 90$ $90 + \angle{H} = 180 \implies \angle{H} = 90$ You did ok, but it seems you make problems like this more difficult than necessary. 
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