MathLabs

Problem 2

If aa, bb, and cc are the sides and α\alpha, β\beta, and γ\gamma are the respective angles of a triangle for which a+b=tan⁡γ2(atan⁡α+btan⁡β)a+b=\tan\frac{\gamma}{2}\left(a\tan\alpha+b\tan\beta\right), prove that the triangle is isosceles.
Step 5 of 6: Solve each factor for α=β\alpha=\beta
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In plain words

Both branches of the product being zero lead to the same conclusion, so the case split does not weaken the result.

cos⁡(α+γ2)=0 ⇒ α=β;acos⁡β=bcos⁡α ⇒ sin⁡(α−β)=0 ⇒ α=β\cos\left(\alpha+\frac{\gamma}{2}\right)=0 \ \Rightarrow\ \alpha=\beta; \qquad a\cos\beta=b\cos\alpha \ \Rightarrow\ \sin(\alpha-\beta)=0 \ \Rightarrow\ \alpha=\beta
Detailed analysis

If the first factor vanishes, α+γ2=π2−β−α2=π2\alpha+\frac{\gamma}{2}=\frac{\pi}{2}-\frac{\beta-\alpha}{2}=\frac{\pi}{2}, forcing α=β\alpha=\beta. If instead acos⁡β=bcos⁡αa\cos\beta=b\cos\alpha, the law of sines gives a=2Rsin⁡αa=2R\sin\alpha, b=2Rsin⁡βb=2R\sin\beta, so sin⁡αcos⁡β=sin⁡βcos⁡α\sin\alpha\cos\beta=\sin\beta\cos\alpha, i.e. sin⁡(α−β)=0\sin(\alpha-\beta)=0; since α,β∈(0,π)\alpha,\beta\in(0,\pi) with α−β∈(−π,π)\alpha-\beta\in(-\pi,\pi), this forces α=β\alpha=\beta.