In 1892, W. W. Rouse Ball published a proof for that question.Here is his proof:
Proof:
Step1: Draw an arbitrary triangle such as ABC with AC>BC. We will prove AC=BC.
Step 2: Draw an angle bisector of angle C.
Step 3: Draw perpendicular bisector of AB.
These two lines (the bisector of angle C, and the perpendicular bisector of AB) cannot be the same line or be parallel, as then the triangle ABC would be isosceles. So they must meet at some point E. This point E must be either inside the triangle, or outside the triangle, or on the line segment AB.
Here we have three cases:
Case 1: E is inside the ABC triangle.
Case 2: E is on the segment AB.
Case 3: E is outside the triangle.
Here I am going to prove case 1. We can prove case 2 and 3 with same approach.
Step 4:Connect vertices A and B with point E.
Step 5: Draw perpendicular segments from E to AC and BC.
Now, since, CE is an angle bisector and EF and EG are perpendiculars from angle bisectors to AB and BC, so EF=EG and CF=CG. Thus, right triangles CFE and CEG are congruent. Also , right triangles ADE and DBE are also congruent. Since DE is a perpendicular to the side AB so AE=BE. And right triangles AEF and BEG are congruent so AF=BG. Thus, AC=BC by adding of equals to equals. Therefore, triangle ABC is isosceles.
The Fawn in the proof: In our classrooms we have learned that every triangle is not isosceles. So the math which we have learned in the classroom is wrong or there is a mistake with Ball's proof. Of course, the math which have learned is not wrong so there is a mistake with the proof. Let use GSP and demonstrate this situation:
As one can see from above picture intersection of angle bisector CE and perpendicular DE is out side the triangle. It never falls inside or on the triangle. Also F is outside the triangle since
CF = CG then BC > AC . This is a contradiction with our choice.
Here is a GSP document to show steps for above picture: