Answer:
ΔAMD ≅ ΔCMD by Side Side Side (SSS) congruency rule
Step-by-step explanation:
Given that ΔABC is an isosceles triangle, with [tex]\overline{BA}[/tex] ≅ [tex]\overline{BC}[/tex]
With BM as the median line from B to AC, we have;
∠ABM = ∠CBM
Also we have;
[tex]\overline{BD}[/tex] ≅ [tex]\overline{BD}[/tex] - Reflexive property
Therefore, ΔABD ≅ ΔCBD Side Angle Side SAS condition of congruency
Therefore;
[tex]\overline{AD}[/tex] ≅ [tex]\overline{CD}[/tex] Corresponding sides of congruent triangles are congruent CPCTC
[tex]\overline{AM}[/tex] ≅ [tex]\overline{CM}[/tex] - AC is bisected by [tex]\overline{BM}[/tex] where [tex]\overline{BM}[/tex] = Median line
[tex]\overline{DM}[/tex] ≅ [tex]\overline{DM}[/tex] - Reflexive property
Therefore, we have;
ΔAMD ≅ ΔCMD - Side Side Side (SSS) congruency rule