<p>nm thats wrong</p>
<p>If X = midpoint....</p>
<p>ABC = 80 (60+20)
BCA = 80 (70+10)
BAC = 20</p>
<p>BXC = 50
BCX = 70
CBX = 60</p>
<p>BDX = 30
DXB = 130
XBD = 20</p>
<p>CXE = 130
ECX = 10
XEC = 40</p>
<p>DAE = 20
AED = 40
EDA = 120</p>
<p>DEX = 100
EXD = 50
XDE = 30 (angle we are trying to find)</p>
<p>78 is my answer..not sure though</p>
<p>Gah! Bernard's checks out too. Which one of the three solutions is right? :confused:</p>
<p>no i screwed up again..its 60</p>
<p>My answer doesn't work if you use alternate angles though</p>
<p>DE and BC aren't parallel, though, are they?</p>
<p>no, they are not!</p>
<p>Bernardolw, I was using ur same method, but dont get how u suddenly out of nowhere figured out AED? didn catch that. xplain plz</p>
<p>If I knew that angle, the rest is just really simple geometry.</p>
<p>what kind of problem is this, by the way? College level mathematics?:rolleyes:</p>
<p>and I agree with upenn_rocks. That 40 degree measure appears out of nowhere.</p>
<p>42? (10 char)</p>
<p>20 or 10-----</p>
<p>Wow. I remember absolutely nothing from geometry.</p>
<p>Edc = 110
.....</p>
<p>how clever... i almost got it... ><</p>
<p>everyone plays a fool</p>
<p>70</p>
<p>i agree...70...i think</p>
<p>it can't be 70. if it is, then you assume that triangle ADE and triangle ABC are similar, which isn't the case because angles EBC and DCB are not equal.</p>
<p>EDC must be an acute angle that is less than 70, though..</p>
<p>Hey guys, I got a solution!!!<br>
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Post this on the MIT forum.</p>
<p>I won't say the answer till someone knows it with 100% certainty... otherwise it's just a big guessing game :P</p>