If D is a point on side BC of a triangle ABC such that AD \(\perp\) BC and BD = DC. Then,
1
\(\Delta\) ABD \(\cong\) \(\Delta\) ACD (By RHS)
2
\(\Delta\) ABD \(\cong\)\(\Delta\) DAC (By RHS)
3
\(\Delta\) ABD \(\cong\) \(\Delta\) ACD (By SAS)
4
\(\Delta\) ABD \(\cong\)\(\Delta\) DAC (By SAS)