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 March 19th, 2019, 05:50 PM #1 Senior Member   Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus 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: 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. 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 Last edited by Chemist116; March 19th, 2019 at 06:06 PM. Reason: added information missing from drawing March 19th, 2019, 10:17 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 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 ... Thanks from topsquark and SenatorArmstrong March 20th, 2019, 06:29 AM   #3
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Quote:
 Originally Posted by skeeter 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 $|EG|^2 + |GF|^2 = |EF|^2$
Bewteefull Mike!!

Trying to make up a similar problem where triangle EFG
is a 3-4-5 right triangle; FG=3, EG=4, EF=5.
Any tips? Thanks.

Last edited by Denis; March 20th, 2019 at 06:49 AM. March 20th, 2019, 03:27 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Duhhhh...works for any right triangle with sides similarly given... With 3-4-5, fits inside isosceles right triangle with equal sides = 7. Thus solution always 90 degrees. March 20th, 2019, 04:22 PM   #5
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Quote:
 Originally Posted by skeeter 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 ...
Pretty slick, Slick! -Dan March 20th, 2019, 05:57 PM   #6
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Quote:
 Originally Posted by skeeter 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 ...
I tried to follow the steps that you mentioned. To reflect the changes I modified the drawing. However I couldn't find a way to justify the angle at $\angle EGF = 90 ^{\circ}$ without using the cosines law as follows:

See figure for more details: $\left(5 \sqrt{3}\right)^{2}= \left(5 \sqrt{2}\right)^{2} + \left(5 \right)^{2} - 2 \left(5^{2} \sqrt{2}\right)\left(\sqrt{3}\right)\cos\angle EGF$

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?. Last edited by Chemist116; March 20th, 2019 at 06:02 PM. March 20th, 2019, 06:43 PM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 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. Last edited by skeeter; March 20th, 2019 at 07:01 PM. March 20th, 2019, 06:46 PM   #8
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Quote:
 Originally Posted by Chemist116 I couldn't find a way to justify the angle at $\angle EGF = 90 ^{\circ}$ without using the cosines law
Triangle EFG is a right triangle...so angle EGF = 90 degrees. March 21st, 2019, 05:15 PM   #9
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Quote:
 Originally Posted by skeeter Pythagoras?
So, next time I get a triangle whose I know its sides should I try to look if it follows Pythagoras theorem?. It looks like guessing, or just trial and error. I felt that using cosines law seemed a bit more rigorus proof.

Quote:
 Originally Posted by skeeter Since angle EGF = 90, then angle EGB and angle FGC are complementary $\implies$ angle A and angle D are complementary $\implies$ angle H = 90.
I'm not sure if what I attempted to do with the drawing which is to be seen below is exactly what you mentioned but I did my best.

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. Therefore from the figure:

$\omega+\phi+\psi=180^{\circ}$

$\psi=180-\left(\omega+\phi\right)=180-90=90^{\circ}$

Again I'm not sure if this was what you meant but so far I cannot find other explanation. Did I do well?.  March 21st, 2019, 06:03 PM   #10
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Quote:
 Originally Posted by Chemist116 So, next time I get a triangle whose I know its sides should I try to look if it follows Pythagoras theorem?. It looks like guessing, or just trial and error. I felt that using cosines law seemed a bit more rigorus proof.
The law of cosines is the general case of the Pythagorean theorem.
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:
 Again I'm not sure if this was what you meant but so far I cannot find other explanation. Did I do well?
The three angles of the large triangle, $\angle{A} + \angle{D} + \angle{H} = 180$

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. Tags angle, coming, find, intersecting, segments, trapezoid Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Math3ka Geometry 4 November 10th, 2015 03:49 AM coder Calculus 3 July 23rd, 2015 12:09 PM 1101 Algebra 4 February 22nd, 2011 12:43 PM tobymac Algebra 1 July 21st, 2009 11:45 PM dai_lo Calculus 1 February 27th, 2008 12:45 PM

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