The triangles are drawn in the picture attached.
Let's call M the intersection point of AB and CD. Since CD is the perpendicular bisector, we know that:
AM ≡ MB (bisector = divides into two equal pieces)
∠AMD ≡ ∠AMC ≡ ∠BMC ≡ ∠BMD (perpendicular = forms 4 angles of 90°)
Considering ΔAMD and ΔBMD, they have also MD in common, therefore, we can use the SAS (side - angle - side) congruency criterium to prove that they are congruent.
Similarly for ΔAMC and ΔBMC.
Therefore, <span>ΔADC ≡ ΔBCD because they are made of congruent triangles.</span>
Hence, with the SAS congruency theorem, we can demonstrate that ΔADC ≡ ΔBCD
Let's call M the intersection point of AB and CD. Since CD is the perpendicular bisector, we know that:
AM ≡ MB (bisector = divides into two equal pieces)
∠AMD ≡ ∠AMC ≡ ∠BMC ≡ ∠BMD (perpendicular = forms 4 angles of 90°)
Considering ΔAMD and ΔBMD, they have also MD in common, therefore, we can use the SAS (side - angle - side) congruency criterium to prove that they are congruent.
Similarly for ΔAMC and ΔBMC.
Therefore, <span>ΔADC ≡ ΔBCD because they are made of congruent triangles.</span>
Hence, with the SAS congruency theorem, we can demonstrate that ΔADC ≡ ΔBCD