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| <math> \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ </math> | | <math> \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ </math> |
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| ==Solution==
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| As a strategy, think of how <math>\angle B + \angle D</math> would be determined, particularly without determining either of the angles individually, since it may not be possible to determine <math>\angle B</math> or <math>\angle D</math> alone. If you see <math>\triangle BFD</math>, then you can see that the problem is solved quickly after determining <math>\angle BFD</math>.
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| But start with <math>\triangle AGF</math>, since that's where most of our information is. Looking at <math>\triangle AGF</math>, since <math>\angle F = \angle G</math>, and <math>\angle A = 20</math>, we can write:
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| <math>\angle A + \angle G + \angle F = 180</math>
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| <math>20 + 2\angle F = 180</math>
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| <math>\angle AFG = 80</math>
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| By noting that <math>\angle AFG</math> and <math>\angle GFD</math> make a straight line, we know
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| <math>\angle AFG + \angle GFD = 180</math>
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| <math>80 + \angle GFD = 180</math>
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| <math>\angle GFD = 100</math>
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| Ignoring all other parts of the figure and looking only at <math>\triangle BFD</math>, you see that <math>\angle B + \angle D + \angle F = 180</math>. But <math>\angle F</math> is the same as <math>\angle GFD</math>. Therefore:
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| <math>\angle B + \angle D + \angle GFD = 180</math>
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| <math>\angle B + \angle D + 100 = 180</math>
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| <math>\angle B + \angle D = 80^\circ</math>, and the answer is thus <math>\boxed{D}</math>
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| == Video Solution == | | == Video Solution == |
Problem
If 
and 
, then 
![[asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,WSW); label("$G$",G,NW);[/asy]](//latex.artofproblemsolving.com/8/2/8/8289a149effe74716d870e23680293e2397d8132.png)
Video Solution
https://www.youtube.com/watch?v=8ntXubG2Iho ~David
See Also
These problems are copyrighted © by the Mathematical Association of America. 
and 
, then 
![[asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,WSW); label("$G$",G,NW);[/asy]](http://latex.artofproblemsolving.com/8/2/8/8289a149effe74716d870e23680293e2397d8132.png)
