# Finding angles of $\triangle ABC$: $CA=CB$, $BD$ is angle bisector and $BD+DC=AB$.

Triangle $$ABC$$ is isosceles ($$CA=CB$$). $$BD$$ is angle bisector of $$\angle B$$. Find angles of triangle $$ABC$$ if $$BD+DC=AB$$.

Actually, I have a solution but I don't like it too much:

If you apply law of sines to $$\triangle BCD$$, you get:

$$\frac{x}{\sin\alpha}=\frac{a-x}{\sin(180^\circ-4\alpha)}\tag{1}$$

Same thing for $$\triangle ABD$$:

$$\frac{a-x}{\sin2\alpha}=\frac{a}{\sin(180^\circ-3\alpha)}\tag{2}$$

Eliminating $$x$$ is easy but you'll have to simplify the resulting equation. After about an hour I got something that looked like a good starting point:

$$2\cos2\alpha(\sin3\alpha-\sin2\alpha)=\sin\alpha\tag{3}$$

But I almost gave up here because if you try to expand multiple angle items like $$\cos2\alpha$$ or $$\sin3\alpha$$, you get a nasty equation that cannot be solved. However, my last attempt succeeded... I replaced $$(\sin3\alpha-\sin2\alpha)$$ with product, cancelled common factors on both sides, then replaced product on the left side with sum and eventually got the following:

$$\cos\frac{9\alpha}2=0\implies \alpha=20^\circ$$

So much work for such a "nice" angle! Considering the simplicity of the result I doubt that there has to be a simpler way to solve this problem without trignonometry.

However, I was not able to find it and so I decided to share this problem with you.

Let $$E \in AB$$ such that $$BD = BE$$ so $$\triangle DEB$$ is isosceles and $$AE=DC$$ (1).
Let $$DS \parallel AB$$ so $$\angle SDB = \angle DBA = \alpha$$.
From here we get $$\triangle DSB$$ is isosceles and $$DS = SB$$ but $$DS \parallel AB$$ imply also $$AD = SB$$ so $$DS = AD$$ (2)
Using (1), (2) we get immediately that $$\triangle AED = \triangle DCS$$ so $$\angle ADE = 2\alpha$$
From here we get $$\angle DEB = 4\alpha$$ (as exterior angle for $$\triangle AED$$) and finally we write for the isosceles triangle $$\triangle DEB$$ the relation $$4\alpha + 4\alpha + \alpha = 180^\circ$$ and we get $$\alpha = 20^\circ$$ etc.