Given isoscelese triangle ABC such that AB = BC and BD bisects angle ABC prove BD is perpendicular to AC.
Any starting points?
Triangles ABD and CBD are congruent by SAS (Side-Angle-Side)
Angles BDA and BDC are contruent by corresponding parts of congruent triangles are congruent (CPCTC)
Angles BDA and BDC are adjacent angles whose non common sides form a straight line (commonly known as a linear pair)
The sum of the measures of the angles of a linear pair is 180 degrees.
Since the two angles are congruent, each angle measures 90 degrees.
If two lines (BD and AC) intersect to form right angles, then they are perpendicular.
Therefore, BD is perpendicular to AC