Triangle ABC is isosceles, equal angles at B and C = 80.Code:`A`

E

D

C B

D is on AC, such that CD = BC.

E is on AB, such that angle BCE = 60.

Find size of angle DEC. No trigonometry allowed! Have fun.

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- Feb 14th 2011, 06:42 PMWilmerGeometry puzzleCode:
`A`

E

D

C B

D is on AC, such that CD = BC.

E is on AB, such that angle BCE = 60.

Find size of angle DEC. No trigonometry allowed! Have fun. - Feb 14th 2011, 10:04 PMalexmahone
- Feb 15th 2011, 02:27 AMWilmer
BUT your "pencil" you "drew" with could be wide enough to be off a degree or so !!

Nice try, Alex.

But now that you have a diagram to play with, go "get the proof" :) - Feb 15th 2011, 12:39 PMWilmer
1st hint: place point F on AB such that CF = CB

- Feb 17th 2011, 09:29 PMWilmer
2nd hint: examine triangle CFD

- Feb 18th 2011, 07:24 PMsimplependulum
Let $\displaystyle F $ be the reflection of $\displaystyle B $ across the foot of perpendicular of $\displaystyle C $ to $\displaystyle AB $ . Then $\displaystyle CF = CB $ , $\displaystyle \angle CFB = 80^o $ . We find that $\displaystyle \angle ECF = \angle CFB - \angle CEF = 80^o - 40^o = \angle CEF$ so $\displaystyle CF = EF $ . At the same time , $\displaystyle \angle DCF = 60^o $ and $\displaystyle CD = CB = CF $ so $\displaystyle \Delta DCF $ is equilateral . We thus have $\displaystyle FC = FD = FE $ . Therefore , $\displaystyle F $ is the circumcentre of $\displaystyle \Delta DCE$ so $\displaystyle \angle CED = \frac{1}{2}~ \angle CFD = \frac{1}{2}~ 60^o = 30^o $

- Feb 19th 2011, 01:32 AMWilmer
Nice. I like this solution (# means angle) :

Triangle BCE: it is given that #ABC = 80 and #BCE = 60; then #BEC = 40

Triangle CDF: this triangle being equilateral, then #CFD = 60;

since #BFC = 80, this leaves #DFE = 180-80-60 = 40

Triangle DEF: #EDF = #DEF, so #CED + 40 = 70, hence #CED = 30