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}`

`\rightarrow tan \alpha =\frac{1}{\sqrt{3}}`

`\rightarrow \alpha = 30`°

Now `\alpha=30`°

In `\triangle PCE`,

`\angle PCE + \angle CPE + \angle PEC =180`°

(Angle sum property)

`\rightarrow 30°+90°+\angle PEC =180°`

`\rightarrow \angle PEC=60°`

As we see to our original figure, `\angle PEC=\angle CED=60°`

Hence `\angle CED=60°`

Thanks!!!