5. On the extension of side $A C$ of triangle $A B C$ beyond point $A$, point $T$ is marked such that $\angle B A C=2 \angle B T C$. Find the area of triangle $A B C$, given that $A B=A C, B T=42$, $A T=29$.
Answer: 420.

Solution. Let $\angle B T A=\alpha$, then by the condition $\angle B A C=2 \alpha$. Triangle $A B C$ is isosceles with base $B C$, so $\angle A B C=\angle A C B=\frac{1}{2}\left(180^{\circ}-2 \alpha\right)=90^{\circ}-\alpha$. By the sum of angles in triangle $T B C$, we get that $\angle T B C=180^{\circ}-\angle T C B-\angle B T C=90^{\circ}$. Moreover, $\angle T B A=\angle T B C-\angle A B C=\alpha$. From the equality of angles, it follows that triangles $A B T$ and $A B C$ are isosceles, $A C=A B=A T$.

Then we get $C T=2 A T=58, B C=\sqrt{C T^{2}-B T^{2}}=40, S_{\triangle B T C}=\frac{1}{2} \cdot B C \cdot B T=840$, $S_{\triangle A B C}=\frac{1}{2} S_{\triangle B C T}=420$.