Find ∠A & ∠C in Triangle Geometry

[Lemoine]

Let △ABC be a triangle. Let AD and CE be its internal bisectors, with D lying on BC and E lying on AB. Given that ∠CED=18° and ∠ADE=24°, how can I find angles ∠A and ∠C without aid of softwares? Angle ∠B is easy to calculate, as
∠CED+∠ADE=∠CAD+∠ACE=∠A+∠C2=180°−∠B2
42°=180°−∠B2
∠B=96°
The other angles are difficult to calculate. I made this construction on Geogebra and got ∠A=12° and ∠C=72°(exactly), but I'm not being able to find them by geometric constructions. A teacher suggested me to drop perpendiculars from E to AD and from D to CE. Let E′ be the reflection of E over AD and D′ be the reflection of D over CE. Both D′ and E′ lie on AC, as AD bisects EE′ and CE bisects DD′.

What I found:
∠CED′=18°
∠ADE′=24°
∠DEE′=∠DE′E=90°−24°=66°
∠EDD′=∠ED′D=90°−18°=72°
∠E′ED′=66°−2⋅18°=30°
∠D′DE′=72°−2⋅24°=24°

 

[Opalg]

Let △ABC be a triangle. Let AD and CE be its internal bisectors, with D lying on BC and E lying on AB. Given that ∠CED=18° and ∠ADE=24°, how can I find angles ∠A and ∠C without aid of softwares? Angle ∠B is easy to calculate, as
∠CED+∠ADE=∠CAD+∠ACE=∠A+∠C2=180°−∠B2
42°=180°−∠B2
∠B=96°
The other angles are difficult to calculate. I made this construction on Geogebra and got ∠A=12° and ∠C=72°(exactly), but I'm not being able to find them by geometric constructions.

I agree that this seems difficult. The only way I could get at it was to use trigonometry.
[TIKZ][scale=1.75]
\coordinate [label=above:{$A$}] (A) at (-6,0) ;
\coordinate [label=right:{$B$}] (B) at (0,0) ;
\coordinate [label=right:{$C$}] (C) at (0.5,3) ;
\coordinate [label=below:{$E$}] (E) at (-2,0) ;
\coordinate [label=right:{$D$}] (D) at (0.25,1.5) ;
\draw (A) -- (B) -- (C) -- (A) -- (D) -- (E) -- (C) ;
\node at (-5.1,0.1) {$\alpha$} ;
\node at (-5.1,0.3) {$\alpha$} ;
\node at (0.3,2.5) {$\gamma$} ;
\node at (0,2.6) {$\gamma$} ;
\node at (-0.17,0.15) {$96^\circ$} ;
\node at (-0.3,1.25) {$24^\circ$} ;
\node at (-1.4,0.5) {$18^\circ$} ;
[/TIKZ]
Suppose that the angles at $A$ and $C$ are $2\alpha$ and $2\gamma$. By the sine rule in $\triangle ABD$, \(\displaystyle \frac{BD}{\sin\alpha} = \frac{AB}{\sin(\alpha+96^\circ)}.\)

By the sine rule in $\triangle BEC$, \(\displaystyle \frac{BE}{\sin\gamma} = \frac{BC}{\sin(\gamma+96^\circ)}.\)

Using the sine rule in $\triangle BDE$ and then in $\triangle ABC$, it follows that $$\frac{\sin(\alpha+24^\circ)}{\sin(\gamma+18^\circ)} = \frac{BD}{BE} = \frac{AB\sin\alpha \sin(\gamma+96^\circ)}{BC\sin\gamma \sin(\alpha+96^\circ)} = \frac{\sin(2\gamma)\sin\alpha \sin(\gamma+96^\circ)}{\sin(2\alpha)\sin\gamma \sin(\alpha+96^\circ)} = \frac{\cos\gamma\sin(\gamma+96^\circ)}{\cos\alpha \sin(\alpha+96^\circ)}.$$
Therefore \(\displaystyle \sin(\alpha+24^\circ)\cos\alpha \sin(\alpha+96^\circ) = \sin(\gamma+18^\circ) \cos\gamma\sin(\gamma+96^\circ).\) Since $\alpha + \gamma = 42^\circ$, we can write this as $$\sin(\alpha+24^\circ)\cos\alpha \sin(\alpha+96^\circ) = \sin(60^\circ - \alpha) \cos(42^\circ - \alpha)\sin(138^\circ - \alpha^\circ).$$ In principle, that equation should determine $\alpha$. In practice, it is not so easy. I found that by messing about with sum-to-product trig formulas, I could reduce it to $$\tan\alpha = \frac{2\cos66^\circ - \sin48^\circ}{\cos48^\circ}.$$ To make progress from there, I had to use the geometry of the regular pentagon.

[TIKZ][scale=0.75]
\coordinate [label=above: $C$] (C) at (90:5cm) ;
\coordinate [label=above right: $D$] (D) at (18:5cm) ;
\coordinate [label=above left: $B$] (B) at (162:5cm) ;
\coordinate [label=below: $A$] (A) at (234:5cm) ;
\coordinate [label=below: $E$] (E) at (306:5cm) ;
\coordinate [label=below: $P$] (P) at (-4.755,-4.045) ;
\coordinate [label=above right: $N$] (N) at (0,-4.045) ;
\draw (A) -- node

{$1$} (B) -- node[above left] {$1$} (C) -- node[above right] {$1$} (D) -- node

{$1$} (E) -- node[below] {$1$} (A) -- (P) -- (B) -- (D) ;
\draw (C) -- (N) ;
\draw (0.3,1.25) node {$K$} ;
\draw (-3.4,-3.75) node {$72^\circ$} ;
\draw (-0.35,4.3) node {$54^\circ$} ;
\draw (0.4,4.3) node {$54^\circ$} ;
[/TIKZ]

In that diagram, $CN$ and $BP$ are perpendicular to $AE$. If the sides of the pentagon have unit length then $AN = \frac12$. But $AN = BK - PA = \sin54^\circ - \cos72^\circ$. So $$\sin54^\circ - \cos72^\circ = \tfrac12,$$ $$\sin(48^\circ + 6^\circ) = \cos72^\circ + \cos60^\circ = 2\cos6^\circ \cos66^\circ,$$ $$\sin48^\circ\cos6^\circ + \cos48^\circ\sin6^\circ = 2\cos6^\circ \cos66^\circ,$$ $$\cos48^\circ\sin6^\circ = \cos6^\circ(2\cos66^\circ - \sin48^\circ),$$ $$\tan6^\circ = \frac{2\cos66^\circ - \sin48^\circ}{\cos48^\circ}.$$ Therefore $\tan\alpha = \tan6^\circ$, so that $\angle A = 12^\circ$ and $\angle C = 72^\circ$, as forecast by Geogebra.​