In a ∆ABC, CA=CB. D and E are points on AB such that AD=DE=EB. If C=120° find ∠CED
In a ∆ABC, CA=CB. D and E are points on AB such that AD=DE=EB. If C=120° find ∠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 △CDP and △CEP: CP=CP (given) ∠CPD=∠CPE=90° DP=PE (as AP=PB and AD=EB. So AP-AD=AD-EB hence DP=PE) ∴...........(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}...