QUESTION
Let the triangle $ABC$ be $\angle A = 110°$ and $\angle B = 40°$. We consider a point $E$ inside the triangle $ABC$ so that $\angle ECB = 10°$ and $\angle EBC=20°$. Show that $CA=CE$.
my drawing
My ideas
We can show that $\angle C= 30$.
I observed that $BE$ is the bisector of $\angle B$ and I thought of the propriety of the bisector points so I drew some perpendicular from $E$ on the $3$ sides of the triangle. We can see that $EY$ and $EX$ are congruent. The fact that we have some $30$ angles makes me think of the theorem of the $30$ angle.
$AZEY$ is an inscribed quadrilateral.
I don't know what to do further. Hope one of you can help me! Thank you!
Best Answer
Construct $\triangle BEF$ such that $\triangle BEF \cong \triangle BEC$ on side $AB$, point $F$ is on the extension of $BA$ through point $A$.
Since $\angle BEF= \angle BEC = 150^{\circ}$ then $\angle CEF=60^{\circ}$
and
Since $EC=EF$ and $\angle CEF=60^{\circ}$ then $\triangle CEF$ is equilateral triangle.
$\angle AFC= \angle BFE + \angle EFC = 10^{\circ}+60{\circ}=70^{\circ}$
$\angle CAF=70^{\circ}$ (sum of $\angle$'s in $\triangle AFC$)
then $CA=CF$ (sides are $=$ opposite $=$ sides)
and
$CA=CE$