Find area of triangle ABD Let ABC be a right-angled triangle $B=90$. Let in a triangle pick a point $D$ inside the triangele such that $AD= 20, DC=15, DB=10$ and $AB=2BC$. What is the area of triangle $ABD$?
Thanks!


i have got three equations
$$5a^2=20^2+15^2-2\cdot 20\cdot 15\cos(2\pi-\alpha-\beta)$$
$$a^2=15^2+10^2-2\cdot 15\cdot 10\cos(\beta)$$
$$4a^2=10^2+20^2-2\cdot 10\cdot20\cos(\alpha)$$
solving this we get $$\alpha=\arctan\left(\frac{1}{2}\right)$$
and $$A_{\Delta ABD}=\frac{1}{2}20\cdot 10\sin\left(\arctan\left(\frac{1}{2}\right)\right)$$
