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

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?

• Can you be explicit about exactly what claim contradicts the congruence in question?
– Him
Dec 4, 2019 at 21:18

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 $$65^\circ$$.