From the data from the figure:

enter image description here

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?

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

1 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 $65^\circ$.

  • $\begingroup$ I didn't know the right name of the congruence. I edited it - Angle-Angle-Side (two angles and the non-included side). $\endgroup$
    – rivan.ares
    Dec 4, 2019 at 21:31
  • $\begingroup$ That's fine, but don't mix them. Either is the included side, or the non included side, but be consistent. In one of the triangles you use the included side, in the other you don't. $\endgroup$
    – Andrei
    Dec 4, 2019 at 21:38

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