Isn't this a contradiction with Angle-Angle-Side congruence?

Question

From the data from the figure:

A) if $\angle CBD = 25^\circ$ , then $\triangle АВС \cong \triangle CDB$

B) if $АС \perp CD$, then $AB = CD$

C) if $AC||BD$, then $BC=AD$

D) $\triangle ABC \not\cong CDB$

I know that the correct answer is $D$ because the hypotenuse of $\triangle BCD$ is a cathetus in $\triangle ABC$. Therefore, $AC>BC$ and the triangles cannot be congruent. Assuming $A$ is the correct answer, we have $BC$ - common side, $\angle BDC=\angle ABC=90^\circ$ and $\angle CBD=\angle ACB=25^\circ$. We have one side and two angles. Isn't this the Angle-Angle-Side congruence?

Answer 1

The Angle-Side-Angle congruence means that two triangles are congruent if one of the sides is the same length and the two adjacent angles are the same. In your case $\angle BDC$ is not next to $BC$, it is opposite. So in one case the angles are $25^\circ$ and $90^\circ$, in the other the angles are $25^\circ$ and $65^\circ$.