Here is a diagram of an isosceles triangle [i]APB[/i] with segment [i]AP[/i] congruent to segment [i]BP[/i].[br][br]Here is a valid proof that the angle bisector of the vertex angle of an isosceles triangle is a line of symmetry.[br][br]Read the proof and annotate the diagram with each piece of information in the proof.[br]

[math]\overline{AP}\cong\overline{BP}[/math] because triangle [i]APB[/i] is isosceles.[br]The angle bisector of [math]\angle APB[/math]intersects [math]\overline{AB}[/math]. Call that point [i]Q.[br][/i]By the definition of an angle bisector, [math]\angle APQ\cong\angle BPQ[/math].[br][math]\overline{PQ}\cong\overline{PQ}[/math] because it is the same segment[br]By SAS [math]\triangle APQ\cong\triangle BPQ[/math].[br][br]Therefore, [math]\overline{AQ}[/math] and [math]\overline{BQ}[/math] are corresponding and congruent segments (CPCTC)[br]and [math]\angle AQP[/math] and [math]\angle BQP[/math] are corresponding and congruent angles. (CPCTC)[br][br]Since [math]\angle AQP[/math] and [math]\angle BQP[/math] are both congruent and supplementary, both angles are right angles.[br]So [math]\overline{PQ}[/math] must be the perpendicular bisector of [math]\overline{AB}[/math].

IM G Unit 2 Lesson 14 from IM Geometry by Illustrative Mathematics, [url=https://im.kendallhunt.com/HS/students/2/2/1/index.html]https://im.kendallhunt.com/HS/students/2/2/14/index.html[/url]. Licensed under the Creative Commons Attribution 4.0 license, [url=https://creativecommons.org/licenses/by/4.0/]https://creativecommons.org/licenses/by/4.0/[/url].