Half Way Points

As shown, $\theta =\alpha - \beta$ and hence $\tan\theta=\frac{-5-3}{1+(-5)(3)}=\frac{4}{7}$ . Now transform the diagram into the following figure, where $B=(-50,0), C=(50,0)$ and $A=(2a,2b)$ which $(2a)^2+(2b)^2=50^2$ .

Note that $D=(-25+a,b), E=(25+a,b)$ as they are midpoints.

Now $\tan\theta=\frac{\frac{b}{-75+a}-\frac{b}{75+a}}{1+\left(\frac{b}{-75+a}\right)\left(\frac{b}{75+a}\right)}=\frac{150b}{a^2+b^2-75^2}=\frac{150b}{25^2-75^2}=-\frac{3b}{100}$ . Since $\tan\theta=\frac{4}{7}$ , then $b=-\frac{400}{21}$ .

Lastly, the area of triangle $ABC$ is $\frac{1}{2}(100)|2b|=\frac{40000}{21}=\frac{p}{q}$ and so $p+q=\boxed{40021}$ .