[b]Proof by transformation:[br][/b]Consider the isosceles triangle ABC with sides [math]AC\cong BC[/math]. Now, take the midpoint of the segment AB to be the point D. The line through DC must be a perpendicular bisector of AB because it lies on the midpoint of AB and the endpoints of the segment AB are equidistant from the point C on the line.[br]Consider the two, new triangles ACD and BCD. Because [math]AC\cong BC[/math], there are exactly two isometries that map A to B and C to C, as shown in the figure above. [br]because DC is a perpendicular bisector of AB, one isometry that maps A to B and C to C is a reflection across the line DC. Because D and C both lie on the line of reflection, they are fixed points and the point A maps to the point B. That is, we have mapped the triangle ACD on top of the triangle BCD. Thus, they are congruent triangles. Because they are congruent triangles, [math]\angle DAC\cong\angle DBC[/math]. Which means [math]\angle BAC\cong\angle ABC[/math] in the triangle ABC. Thus, the base angles of the isosceles triangle are congruent.