In the following figure, triangle $ABC$ is inscribed in circle $C$ and $AD$ is the bisector of $\angle A$.Also it's known that: $\frac{AB}{BC}=\frac{AB}{AD}$.Prove that: $\angle CBA = \angle DAB$ enter image description here

I tried as follows:
It's obvious that $\angle DBC=\angle CAD$ , so $\angle DBC=\angle DAB$.Now it remains to show that: $\angle DBC=\angle CBA$.Maybe triangle $ABC$ and $ABD$ are equall(they have two equal sides and $\angle ADB=\angle ACB$) but I don't see another equal pair of angles between them!

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    $\begingroup$ From the first equation, you have $BC=AD$. Then, you have $CD // AB$. $\endgroup$ – GAVD Oct 26 '17 at 9:10
  • $\begingroup$ @GAVD Elegant point! I believe if we join $C$ and $D$ it's even not required to use parallelism of $CD$ and $AB$ $\endgroup$ – Hamid Reza Ebrahimi Oct 26 '17 at 9:27

But this is easy. Since $AD=BC$ we have $\angle BAC = \angle ABD$. Since $\angle CAD = \angle CBD$ we have:

$$ \angle CBA = \angle DBA -\angle DBC = \angle BAC -\angle CAD = \angle DAB $$

And we don't need angle bisector.

  • $\begingroup$ I found another reasoning: Join $C$ and $D$,now triangle $BDC$ is isosceles and $\angle DCB=\angle BDC$.On the other hand,$\angle DCB=\angle DAB$ then.... $\endgroup$ – Hamid Reza Ebrahimi Oct 26 '17 at 10:26

$\angle ACB= \angle ADB$ since they are angles in the same segment. $BC=AD$ by your ratio condition. Finally since $C,D$ are on the same side of $AB$, $\angle CAB$ and $\angle DBA$ must both be acute or both obtuse. So $\triangle ABC \cong \triangle BAD$, in particular, $\angle CBA=\angle DAB$.


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