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]]>Thanks, for providing more. I started to understand the proof when I discovered it was harder to draw a decent diagram than to understand it.
]]>That is brief! I think a few added details plus diagram below will make it clear .
Draw perpendicular bisector of AE and place Q on this line so that AQE = 20
This triangle will be isosceles so the bottom angles are both 80.
The base AE = BC so AQE and ABC are congruent triangles.
Hence AB = AC = AQ = EQ
So centre A, radius AC also goes through Q (and B)
QAE = 80 => QAC = 60 making AQC an equilateral triangle.
AQC - AQE = 60 - 20 = 40.
Then in the isosceles triangle QEC the angle QEC = 70.
Add QEA = 80 and you're there.
CAQ = EAQ - EAC = 80° - 20° = 60°
ΔACQ is equilateral. So:
ΔCQE, EQ = CQ and CQE = 60° - 20° = 40°
CQ = EQ = AQ.
CEQ = (180° - 40°) / 2 = 70°
AEC = 80° + 70° = 150°.
]]>Looks good from here! Good work!
If you've got a quicker way I'd be glad to see it.
I am looking at a short proof right now, but I do not undertsand it yet. As soon as I can at least follow it, I will post it for you. Please be patient, If I can't get it I will post it anyway because you might follow it and then you can explain it to me.
]]>But now I think I've cracked it. It's a long proof so apologies. If you've got a quicker way I'd be glad to see it.
I moved my question to here where everyone could see it.
http://www.mathisfunforum.com/viewtopic … 26#p146326
post # 136
]]>Here's the diagram I meant to post:
]]>Put it betwen the hide tags.
[ and then "hide" and then ] put your proof here. Then [ , then "\hide" then ]
]]>The diagram shows a triangle ABC
with ABC = 80 and ACB = 80
D lies on AC so that DBC = 60
and E lies on AB so that ECB = 50.
To find (by Euclidean geometry) x = EDB
Then bobbym gave me a new puzzle to try.
Given: A triangle ABC with A = 20 and B = C = 80.
E lies on AB so that AE = BC.
Find angle AEC.
It's taken me a while but I've finally got a solution that I think stands up OK. As soon as I've figured how to hide it, I'll post it.
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