# In the following figure, $AD$ is the bisector of $\angle A$.Prove that: $\angle CBA = \angle DAB$

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

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!

• From the first equation, you have $BC=AD$. Then, you have $CD // AB$.
– GAVD
Oct 26, 2017 at 9:10
• @GAVD Elegant point! I believe if we join $C$ and $D$ it's even not required to use parallelism of $CD$ and $AB$ Oct 26, 2017 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.

• 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.... Oct 26, 2017 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, $\angle CBA=\angle DAB$.