# Geometry puzzle

• Feb 14th 2011, 06:42 PM
Wilmer
Geometry puzzle
Code:

        A               E       D           C          B
Triangle ABC is isosceles, equal angles at B and C = 80.

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 PM
alexmahone
Quote:

Originally Posted by Wilmer
Code:

        A               E       D           C          B
Triangle ABC is isosceles, equal angles at B and C = 80.

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.

DEC=30

Full solution:
Spoiler:
I drew a scale diagram and measured it. (Giggle)
• Feb 15th 2011, 02:27 AM
Wilmer
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 PM
Wilmer
1st hint: place point F on AB such that CF = CB
• Feb 17th 2011, 09:29 PM
Wilmer
2nd hint: examine triangle CFD
• Feb 18th 2011, 07:24 PM
simplependulum
Let $F$ be the reflection of $B$ across the foot of perpendicular of $C$ to $AB$ . Then $CF = CB$ , $\angle CFB = 80^o$ . We find that $\angle ECF = \angle CFB - \angle CEF = 80^o - 40^o = \angle CEF$ so $CF = EF$ . At the same time , $\angle DCF = 60^o$ and $CD = CB = CF$ so $\Delta DCF$ is equilateral . We thus have $FC = FD = FE$ . Therefore , $F$ is the circumcentre of $\Delta DCE$ so $\angle CED = \frac{1}{2}~ \angle CFD = \frac{1}{2}~ 60^o = 30^o$
• Feb 19th 2011, 01:32 AM
Wilmer
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