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

enter image description here

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.

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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. enter image description here

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  • $\begingroup$ Well done, indeed! $\endgroup$ – Oldboy Jan 15 at 19:28

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