How to find the area of triangle ABC if BP is an angle bisector?
2026-05-16 11:15:38.1778930138
How to find the area of triangle ABC?
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$$AB=x;\;BC=y$$ $$\begin{cases} \frac{x}{1} =\frac{y}{2}\to y=2x\\ x^2+y^2=9\\ \end{cases} $$ $$x^2+(2x)^2=9$$ $$x=\frac{3}{\sqrt 5};\;y=\frac{6}{\sqrt{5}}$$ $$Area=\frac{1}{2}xy=\frac{9}{5}$$