@yangtheary I don't see what exactly you want to do with Maple for this question. But I follow my guess. Since you didn't specify which two edges of your triangle are equal, I check all three possibilities.

1- **AC=BC**: In this case **D=C** and we have **ABD=ABC** and since **ABC=BAC**, the answer is **20** degrees.

2- **AB=BC**: Since **AD=BC**, the triangle **ADB** also becomes isosceles. Then **2(?)+20=180** which implies **ABD=?=80**.

3- **AB=AC**: In this case, note that the triangle **ABC** can be uniquely determined by the assumptions of your question plus length of **AD**. And two cases for two different values of **AD** are geometrically similar (one is a scaled version of the other). Therefore the angles are the same and only size of edges are changed (by a multiplying to a fixed ratio). Now I fixed **AD=1**. Without loss of generality (at most with a geometric translation and rotation which none of them change the size of angles and edges) assume that **A=(0,0)** and **AC** is on the positive side of the **x**-axis. Then it's trivial that **D=(1,0)** and the **AB** edge is on the line **y=tan(20)*x**. The point **C** is **(xC,0)** which we yet don't know the value of **xC**. The line which **BC** is on it, is **y=-tan(80)*(x-xC)**. The point **B** is the unique point on the intersection of the two lines and with the distance **1** from **C**. Therefore we need to solve the following system of equations. Here I use Maple. Just note that **B** is on these two lines, by considering one of them, say the first one, we have **B=(xB,tan(20)*xB)**. Therefore we have two equations and two unknowns.

solve({tan(20*Pi/180)*xB=-tan(80*Pi/180)*(xB-xC),(xC-xB)^2+(tan(20*Pi/180)*xB-0)^2=1,xB>0,xC>0},{xB,xC});

Now we have the coordinates of both **B** and **C**. But knowing **B** is enough for us. Again using Maple we compute the arctan of **ABD**. To this end, just choose an arbitrary point on one of the two edges of this angle, say **D** from **BD**, then project it on the other edge. Call this new point **M**. Then **tan(ABD)=(distance of M and D)/(distance of M and AB)**. The wollowing Maple computations is related to all these steps. To find **M**, we first found distance of **M** from the line of **AB**. Then solved an equation determining the unique point on line of **AB** with the found distance from **M**.

pointA:=[0,0]:
pointD:=[1,0]:
sol1:=solve({tan(20*Pi/180)*xB=-tan(80*Pi/180)*(xB-xC),(xB-xC)^2+(tan(20*Pi/180)*xB-0)^2=1,xB>0,xC>0},{xB,xC});
pointB:=eval([xB,tan(20*Pi/180)*xB],sol1);
pointC:=eval([xC,0],sol1);
d:=abs((-tan(20*Pi/180))*pointD[1]+(1)*pointD[2])/sqrt((-tan(20*Pi/180))^2+(1)^2);
sol2:=solve({(xM-pointD[1])^2+(tan(20*Pi/180)*xM-pointD[2])^2=d^2,xM>0},xM);
pointM:=eval([xM,tan(20*Pi/180)*xM],sol2);
theta:=arctan(d/sqrt((pointM[2]-pointB[2])^2+(pointM[1]-pointB[1])^2))*180/Pi;
evalf(theta);
P1:=plots:-pointplot([pointA,pointB,pointC,pointD,pointM],color=black,symbol=solidcircle,symbolsize=20):
P2:=plots:-textplot([[op(pointA),"A"],[op(pointB),"B"],[op(pointC),"C"],[op(pointD),"D"],[op(pointM),"M"]],align={above,right},color=red,font=["times",20,bold]):
P3:=plot([pointA,pointB,pointC,pointA],color=blue,thickness=8):
P4:=plot([pointB,pointD],color=blue,thickness=8):
P5:=plot([pointD,pointM],color=blue,thickness=8):
plots:-display(P3,P4,P5,P1,P2,view=[0..3,0..1.2],size=[700,280],scaling=constrained);

So the answer in the case of **AC=AB** is (almost) **10** degrees.

I also put the plot of my above code below.