Triangle geometry: Let △ABC be a triangle. Let AD and CE be its internal bisectors, w

[Lemoine]

Triangle geometry: Let △ABC be a triangle. Let AD and CE be its internal bisectors, w

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°

 

[orecchionebruno]

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°

Could you post a drawing?

 

[Lemoine]

As I understand triangleABC:

                         A







C                    B

D is on BC, E is on AB.
Changing your D' to F and your E' to G.
So the perpendiculars are CF and EG.\
Let H be the intersection point of bisectors AD and CE.

We now have right triangles DFH and EGH.
These are SIMILAR right triangles: both 90-48-42

That should be enough for you to wrap up this mess

How can I know these triangles are right? How can I know their angles?