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!

 A: Let $BC=a$ and $AB=2a$. 
With $B$ as origin let the co-ordinates of $D$ be $(x,y)$. The required area of triangle $ABD$ is $ax$.
Applying Pythagoras' Theorem $$10^2=x^2+y^2 \text{  ... (1)}$$ $$15^2=(a-x)^2+y^2 \text{ ... (2)}$$ $$20^2=x^2+(2a-y)^2 \text{  ... (3)}$$
Subtract (1) from (2) and (3) in turn : $$125=a(a-2x)\implies 2ax=a^2-125$$ $$300=2a(2a-2y) \implies 2ay=2a^2-150$$
Substitute into (1) : $$100(2a)^2=(2ax)^2+(2ay)^2=(a^2-125)^2+(2a^2-150)^2$$ $$400a^2=a^4-250a^2+125^2+4a^4-600a^2+150^2$$ $$0=5a^4-1250a^2+38125$$ $$0=a^4-250a^2+7625$$ $$a^2=\frac12(250\pm \sqrt{250^2-4.7625})=125\pm 40\sqrt{5}$$
We reject the smaller value which corresponds to $D$ outside of the triangle.
$$2ax=a^2-125=40\sqrt{5}$$ The area of triangle $ABD$ is $$ax=20\sqrt{5} \approx 44.72$$
A: 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 $$A_{\Delta ABD}=\frac{1}{2}20\cdot 10\sin\left(\arctan\left(\frac{1}{2}\right)\right)$$
