As C is the mid-point of BD.
Therefore, BC=DC ...(i)
As AB ⊥ BD and ED ⊥ BD
So, ∠ABC= 90° and ∠EDC= 90°
Therefore, ∠ABC=∠EDC= 90° ...(ii)
As two line segments, AE and BD intersect at point C, so the vertically opposite angles are equal.
Therefore, ∠BCA=∠DCE ...(iii)
Now, in ΔABC and ΔEDC,
Angle, ∠BCA=∠DCE [ from equation (iii)]
Side, BC=DC [ from equation (i)]
Angle, ∠ABC=∠EDC [ from equation (ii)]
So, by ASA property of congruency, ΔABC and ΔEDC are congruent
Hence, ΔABC ≅ ΔEDC.
