In a ∆ABC, CA=CB. D and E are points on AB such that AD=DE=EB. If C=120° find `\angle CED`
In a ∆ABC, CA=CB. D and E are points on AB such that AD=DE=EB. If C=120° find `\angle CED` First the figure is CA=CB and AD=DE=EB: Now, let's construct a line segment CP perpendicular to AB to C: Now, by the property of isoscleles triangle, its altitude is its median. So CP divided angle C in two halfs and also bisects AB. So AP=PB Now, in `\triangle CDP` and `\triangle CEP`: CP=CP (given) `\angle CPD `=`\angle CPE`=90° DP=PE (as AP=PB and AD=EB. So AP-AD=AD-EB hence DP=PE) `\therefore \triangle CDP \cong \triangle CEP`...........(i) But DP+PE=DE Now let AD=DE=EB=2x Then DP=PE=x Also, `\angle PCA =\angle PCB =60`° `also ~ \angle PCD=\angle PCE=\alpha`......from(i) Now, let's focus on `\triangle PCB` as follows: Here, `tanC=\frac{PB}{PC}` `\rightarrow tan60`°=`\frac{3x}{PC}` `\rightarrow \frac{3x}{\sqrt{3}}=PC` Now in `\triangle PCE=` `\rightarrow tan \alpha =\frac{PE}{PC} ` `\rightarrow tan \alpha =\frac{x}{\frac{3x}{\sqrt{3}}}` `\rightarrow tan \alpha =\frac{\sqrt{3} x}{3x}`