Finding distance of a vertices of a rectangle from another point in it when distances to other three vertices is given.

Question

Let $X$ be a point inside a rectangle $ABCD$. If $XA=a$, $XB=b$, $XC=c$. Find $XD$.

I tried by using cosine rule on all the triangle formed and equating what I got. Then I tried to find the relations of sines of interior angles that I know will sum up to $360°$ by adding opposite triangles areas and equating them. I tried to then use the $sin(a+b)$ by using all the sines and cosine relations and trying to get them equate to $sin(360°)$ that is zero. But I ended up with a mess and I am stuck.

A solution without trignometry if possible would be better. Hints are always appreciated.

Answer 1

By Pythagoras Theorem, $XD^2=XF^2+XH^2$... (1) $a^2=XH^2+XE^2$... (2) $b^2=XE^2+XG^2$... (3) $c^2=XG^2+XF^2$... (4)

(2)+(4)-(3) gives, $XF^2+XG^2=c^2+b^2-a^2$ By (1), $XD^2=c^2+a^2-b^2$