Given:Ab congruent to CBCD bisects AbProve:triangle ADC is congruent to triangle to BDC

Given triangle ABC, (as shown in the diagram attached) the sides AC and CB are congruent.
Also line CD bisects line AB.
Therefore, line
[tex]\begin{gathered} AD\cong BD \\ Angle\text{ bisector of a triangle} \\ An\text{ angle bisector of a triangle divides the opposite side into two} \\ \text{segments that are prportional }to\text{ the other two sides of the triangle} \\ \text{Hence, if CD bisects line AB, and AC}\cong BC \\ \text{Then AB}\cong BD \end{gathered}[/tex][tex]\begin{gathered} \angle ACD=\angle BCD \\ \text{Angle bisector} \\ \text{If the line CD bisects line AB, and }AD\cong BD \\ \text{Then }\angle ACD\cong\angle BCD \end{gathered}[/tex]Therefore, in both triangles, we have;
[tex]AC\cong CB\text{ (Given)}[/tex][tex]AD\cong BD\text{ (Angle bisector)}[/tex][tex]undefined[/tex]