# Figuring out an angle in an isosceles triangle

A problem from BdMO 2013:

Let $ABC$ be an isoscles triangle with $AB=AC$.The bisector of $\angle B$ meets $AC$ at $D$.Given that $BC=BD+AD$,we need to figure out $\angle A$.

If we consider $\angle B=\angle C=\angle 2x$,then after bisecting angle B,we get a triangle with angles equal to $x$,$2x$,and $180-3x$.But that does not get us any further except that $\angle ADB=3x$.I also tried extending BD to $A'$ such that $A'D=AD$ but that does not help at all.Finally,I tried to utilize the Angle Bisector Theorem but that yielded nothing good as well.A prod in the correct direction would be appreciated.

NOTE: I am looking for a hint,not the whole solution.

• Here $A=C$ and not $B=C$ Dec 5, 2013 at 13:26
• And moreover $BC=BD+AD$ is impossible since $AD=DC$ and triangle $BDC$ would be degenerate! Dec 5, 2013 at 13:27

We take $E$ on $BC$ such that $BE=BD$; so $EC=BC-BE=BC-BD=AD$.
• Now we have a theorem that $\frac{AB}{BC}=\frac{AD}{DC}$; so in $\bigtriangleup CED$ and $\bigtriangleup CAB$ we have a common angle $\frac{CE}{CD}=\frac{AD}{CD}=\frac{AB}{CB}=\frac{CA}{CB}$.
• So we get $\Delta CED~\Delta CAB$ so we have $\angle CDE=\angle DCE=\angle ABC=2x$ [let]
• Hence $\angle BDE=\angle BED=4x$, so $9x=180$, or $x=20$.
• Thus $\angle A=180-4x=100$ • +1.The theorem $AB/BC=AD/DC$ is called angle bisector theoren,just for reference.Btw,I am perplexed at your idea of a hint. Dec 5, 2013 at 18:18