ΔABC and ΔADB are on the common base AB and on the same side of AB. DA ⊥ AB,…

Two right triangles are congruent if the hypotenuse and one corresponding leg of one triangle equal the hypotenuse and the corresponding leg of the other -- this is the RHS (Right angle-Hypotenuse-Side) congruence criterion. To state the congruence correctly, the vertex correspondence must pair each triangle's right-angle vertex with the other's right-angle vertex, and pair equal sides with equal sides.

Applying the criterion to this figure:

1. DA ⊥ AB and CB ⊥ AB, so ∠DAB = 90° and ∠CBA = 90°: the right angle of ΔABC is at B, and the right angle of the second triangle is at A.

2. AC = BD (given): the hypotenuse of ΔABC (opposite the right angle at B) equals the hypotenuse of the second triangle (opposite the right angle at A).

3. AB is common to both triangles, and it is the side adjacent to each right angle (not the hypotenuse in either triangle).

4. By RHS, the right-angle vertices must correspond (B with A), the hypotenuses must correspond (AC with BD), and the common side must correspond to itself (AB with BA). This forces the vertex order A↔B, B↔A, C↔D.

5. Writing the triangles in that matched order gives ΔABC ≅ ΔBAD.

Independent check by placing coordinates: let A = (0, 0) and B = (b, 0). Since DA ⊥ AB, D lies directly above A, so D = (0, d); since CB ⊥ AB, C lies directly above B, so C = (b, e). The condition AC = BD gives √(b²+e²) = √(b²+d²), so e = d. Comparing the three side-pairs under the correspondence A↔B, B↔A, C↔D:

All three corresponding side-pairs match, confirming ΔABC ≅ ΔBAD independently of the RHS argument above.