# Find the angle in an isosceles triangle

Let triangle $$\Delta ABC$$ have $$AB=AC$$. Then we draw the angle bisector from $$B$$ to $$AC$$ intersecting at $$D$$. Find the angle $$\angle BAC$$ if $$BC=AD+BD$$.

My attempts:

I know that the answer is 100° but I couldn't prove that if you extended $$AD$$ to a point $$E$$ so it is equal to $$BC$$, then the angle $$\angle DCE$$ is the same as $$\angle ACB$$.

• I solved your problem. If you want to see my solution show us your trying. – Michael Rozenberg Mar 28 at 10:22
• I know that the answer is 100° but I couldn't prove that if you extended AD to a point E so it is equal to BC the angle DCE is the same as ACB. – AJMC2002 Mar 28 at 17:18
• I posted. See now. – Michael Rozenberg Mar 28 at 17:26

In the standard notation we obtain: $$\frac{AD}{DC}=\frac{AB}{BC}=\frac{c}{a}$$ and $$AD+DC=AC=b,$$ which gives $$AD=\frac{bc}{a+c},$$ $$DC=\frac{ab}{a+c}$$ and $$BD^2=AB\cdot BC-AD\cdot DC=ac-\frac{b^2ac}{(a+c)^2}=ab-\frac{ab^3}{(a+b)^2},$$ which gives $$BD=\frac{a\sqrt{b(a+2b)}}{a+b}.$$ Id est, by the given we obtain: $$a=\frac{b^2}{a+b}+\frac{a\sqrt{b(a+2b)}}{a+b}$$ or $$a\sqrt{b(a+2b)}=a^2+ab-b^2.$$ Now, $$\sin\frac{\alpha}{2}=\frac{\frac{a}{2}}{b}=\frac{a}{2b}.$$

Let $$\sin\frac{\alpha}{2}=x.$$

Thus, $$a=2xb$$ and we obtain: $$2x\sqrt{2(x+1)}=4x^2+2x-1$$ or $$8x^2(x+1)=(4x^2+2x-1)^2,$$ where $$x>0$$ and $$4x^2+2x-1>0.$$

We obtain: $$16x^4+8x^3-12x^2-4x+1=0$$ or $$8x^3(2x+1)-12x^2-6x+2x+1=0$$ or $$(2x+1)(8x^3-6x+1)=0$$ or $$3x-4x^3=\frac{1}{2}$$ or $$\sin\frac{3\alpha}{2}=\frac{1}{2},$$ which gives $$\frac{3\alpha}{2}=30^{\circ},$$ which is impossible because $$4x^2+2x-1>0,$$ or $$\frac{3\alpha}{2}=150^{\circ},$$ which gives $$\alpha=100^{\circ}$$ and $$\beta=\gamma=40^{\circ}.$$

The fact that $$BD^2=AB\cdot BD-AD\cdot DC$$ we can prove by the following reasoning.

Let $$\Phi$$ be a circumcircle of $$\Delta ABC$$ and $$AD\cap\Phi=\{A,E\}$$.

Thus, $$\measuredangle BAC=\measuredangle BEC$$ and $$\measuredangle ABD=\measuredangle EBD,$$ which gives $$\Delta ABD\sim\Delta EBC.$$

Hence, $$\frac{AB}{BE}=\frac{BD}{BC}$$ or $$AB\cdot BC=BD\cdot BE$$ or $$AB\cdot BC=BD(BD+DE)$$ or $$BD^2=AB\cdot BC-BD\cdot DE$$ or $$BD^2=AB\cdot BC-AD\cdot DC.$$

• Could you elaborate on how did you get BD²=AB•BC-AD•DC? Thanks – AJMC2002 Mar 29 at 5:39
• @user658009 I added something. See now. – Michael Rozenberg Mar 29 at 8:13