Consider circle A in the diagram below where m angle DAF = 44° and triangle CGD is isosceles. G is the vertex angle of isosceles triangle CGD.
The m angle CED = (44°, 136°, 68°, or 56°) and the m angle CGD = (44°, 136°, 68°, or 56°) the m angle CDG = (44°, 136°, 68°, or 56°)

Step-by-step explanation:

Check line CAF, it is a straight line, then the sum of angle on a straight line is 180°

Therefore,

<CAD + < DAF = 180°

Given that <DAF = 44°

<CAD + 44° = 180°

<CAD  = 180° - 44°

<CAD = 136°

From circle theorem

Angle subtended at the center of the circle is twice any inscribe angle subtended by the same arc.

Then, looking at arc CD, we notice that CAD is at the centre and CGD is at the circumference and they both form the same arc.

Then, using this theorem

< CAD = 2 × < CGD

We already < CAD = 136

136 = 2 ×  < CGD

Divide both side by 2

< CGD = 136/2

< CGD = 68°

Also, Angel in the same segment are equal, so subtended angle by same arc are equal

So, <CED is in the same arc as  <CD,

Then, <CED = <CGD

<CED = 68°

Now, to get <CDG

Since ∆CGD is isosceles, it implies than two sides are equal, and two angles are equal and since G form the vertex, then,

|CG| = |DG| and <GCD = <CDG =z

So, sum of angle in a triangle is 180°

Then, applying this to ∆CGD

<GCD + <CDG + <CGD = 180°

We already got <CGD = 68°

<GCD + <CDG + 68° = 180°

<GCD + <CDG   = 180°  - 68°

<GCD + <CDG  = 112°

Since, <GCD = <CDG =z

Then, z + z = 112°

2z = 112°

Divide both side by 2

z = 112/2

z = 56°

So,

<CDG = 56°

Answer

<CED = <CGD = 68°

<CDG = 56°