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.
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 $
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