Welcome to the forum. Please start a new thread in "Help Me" for your question.

]]>therefore x > exp(x - 1) = exp(x) / e

x .1 means for all x there must exist p>0 such that x=1+p and substituting this for x we see that

1 + p > exp(1+p)/e = exp(p).e/e = exp(p)

Our assumption produces then 1+p > exp(p) and this is false if p=2,3,4,.......

So assumption is false, proving the result.]]>

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.

AQ = AC.

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.

]]>