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No need to use coordinates of points just sides are enough.

We do not have to solve for s and ag, we already have formulas for them.

Why I used /. Abs -> Identity? Because in Wolfram they probably do not know that Heron's formula should be without Abs.

Solve provides four solutions, two of them evidently negative.

One of other two solutions is wrong on back substitution for reason unknown to me (without any warning message from Solve). So only the second solution is correct.

Solve[{s == Area[SSSTriangle[a, b, c]], ag == TriangleMeasurement[ SSSTriangle[a, b, c], {"InteriorAngle", 2}], Area[SSSTriangle[a, b, c]] == Area[SSSTriangle[d, b, jlcd]] + Area[SSSTriangle[a, c - d, jlcd]]} /. Abs -> Identity, jlcd, {ag, s}] %[[2]] 

No need to use coordinates of points just sides are enough.

We do not have to solve for s and ag, we already have formulas for them.

Why I used /. Abs -> Identity? Because in Wolfram they probably do not know that Heron's formula should be without Abs.

Solve provides four solutions, two of them evidently negative.

One of other two solutions is wrong on back substitution for reason unknown to me (without any warning message from Solve). So only the second solution is correct.

Solve[{s == Area[SSSTriangle[a, b, c]], ag == TriangleMeasurement[ SSSTriangle[a, b, c], {"InteriorAngle", 2}], Area[SSSTriangle[a, b, c]] == Area[SSSTriangle[d, b, jlcd]] + Area[SSSTriangle[a, c - d, jlcd]]} /. Abs -> Identity, jlcd, {ag, s}] %[[2]] 

Update:

Omitting two unnecessary equations provides correctly output with warning.

Solve[Area[SSSTriangle[a, b, c]] == Area[SSSTriangle[d, b, jlcd]] + Area[SSSTriangle[a, c - d, jlcd]] /. Abs -> Identity, jlcd] 

No need to use coordinates of points just sides are enough.

We do not have to solve for s and ag, we already have formulas for them.

Why I used /. Abs -> Identity? Because in Wolfram they probably do not know that Heron's formula should be without Abs.

Solve provides four solutions, two of them evidently negative so we pick up only other two.

Why there are two solutions? Because the triangle sides can be ordered a, b, c or b, a, c.

Solve[{s == Area[SSSTriangle[a, b, c]], ag == TriangleMeasurement[ SSSTriangle[a, b, c], {"InteriorAngle", 2}], Area[SSSTriangle[a, b, c]] == Area[SSSTriangle[d, b, jlcd]] + Area[SSSTriangle[a, c - d, jlcd]]} /. Abs -> Identity, jlcd, {ag, s}] %[[{2, 4}]] 

No need to use coordinates of points just sides are enough.

We do not have to solve for s and ag, we already have formulas for them.

Why I used /. Abs -> Identity? Because in Wolfram they probably do not know that Heron's formula should be without Abs.

Solve provides four solutions, two of them evidently negative.

One of other two solutions is wrong on back substitution for reason unknown to me (without any warning message from Solve). So only the second solution is correct.

Solve[{s == Area[SSSTriangle[a, b, c]], ag == TriangleMeasurement[ SSSTriangle[a, b, c], {"InteriorAngle", 2}], Area[SSSTriangle[a, b, c]] == Area[SSSTriangle[d, b, jlcd]] + Area[SSSTriangle[a, c - d, jlcd]]} /. Abs -> Identity, jlcd, {ag, s}] %[[2]] 

No need to use coordinates of points just sides are enough.

We do not have to solve for s and ag, we already have formulas for them.

Why I used /. Abs -> Identity? Because in Wolfram they probably do not know that Heron's formula should be without Abs.

Solve provides four solutions, two of them evidently negative so we pick up only other two.

Why there are two solutions? Because the triangle sides can be ordered a, b, c or b, a, c.

(* s==Area[SSSTriangle[a,b,c]]; ag==TriangleMeasurement[SSSTriangle[a,b,c],{"InteriorAngle",2}]; *)  Solve[Area[SSSTriangle[a, b, c]] == Area[SSSTriangle[d, b, jlcd]] + Area[SSSTriangle[a, c - d, jlcd]] /.  Abs -> Identity, jlcd, {ag, s}] %[[{2, 4}]] 

No need to use coordinates of points just sides are enough.

We do not have to solve for s and ag, we already have formulas for them.

Why I used /. Abs -> Identity? Because in Wolfram they probably do not know that Heron's formula should be without Abs.

Solve provides four solutions, two of them evidently negative so we pick up only other two.

Why there are two solutions? Because the triangle sides can be ordered a, b, c or b, a, c.

Solve[{s == Area[SSSTriangle[a, b, c]],   ag == TriangleMeasurement[ SSSTriangle[a, b, c], {"InteriorAngle", 2}],  Area[SSSTriangle[a, b, c]] ==   Area[SSSTriangle[d, b, jlcd]] +  Area[SSSTriangle[a, c - d, jlcd]]} /. Abs -> Identity, jlcd, {ag,  s}] %[[{2, 4}]]