Let the triangle $ABC$ be $\angle A = 110°$ and $\angle B = 40°$. We consider a point $E$ . Show that $CA=CE$

Question

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!

Answer 1

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

$CA=CE$

Answer 2

In $\triangle ABC$ \begin{align*} \frac{AC}{\sin40^\circ}&=\frac{BC}{\sin110^\circ}\\ AC&=\frac{\sin40^\circ}{\sin70^\circ}BC \end{align*} In $\triangle EBC$ \begin{align*} \frac{EC}{\sin20^\circ}&=\frac{BC}{\sin30^\circ}\\ EC&=2\sin20^\circ BC \end{align*} Now we have to prove $$2\sin20^\circ=\frac{\sin40^\circ}{\sin70^\circ}$$ It can be proved by using product-to-sum formula. \begin{align*} \iff&&2\sin20^\circ&=\frac{\sin40^\circ}{\sin70^\circ}\\ \iff&&\sin20^\circ\sin70^\circ&=\sin40^\circ\\ \iff&&\cos(70^\circ-20^\circ)-\cos(70^\circ+20^\circ)&=\sin40^\circ\\ \iff&&\cos50^\circ&=\sin40 \end{align*}

Answer 3

One more "adventitious angles" problem...

This one is based on the following trigonometric relationship : $$\frac{\sin 40°}{\sin 110°}=\frac{\sin 20°}{\sin 150°}\tag{1}$$

Indeed, taking advantage of the result of the angle chasing you you have done, let us apply sine law to triangles $ABC$ and $EBC$ (we can assume WLOG that $BC=1$) :

$$\frac{AC}{\sin 40°}=\frac{1}{\sin 110°} \ \ \text{and} \ \ \frac{CE}{\sin 20°}=\frac{1}{\sin 150°} $$

Taking (1) into account, we derive $AC=CE$.

Edit : proof of (1)

(1) is equivalent to :

$$\frac{\sin 40°}{\cos 20°}=\frac{\sin 20°}{\cos 60°},$$

itself equivalent to

$$\frac{2 \sin 20° \cos 20°}{\cos 20°}=\frac{\sin 20°}{\tfrac12}$$

which is true.